PHY472

Advanced Quantum Mechanics

Pieter Kok, The University of Sheffield.

August 2013

PHY472: Lecture Topics

1. Linear vector spaces

2. Operators and the spectral decomposition

3. Observables, projectors and time evolution

4. Tensor product spaces

5. The postulates of quantum mechanics I

6. The postulates of quantum mechanics II

7. The Schrödinger and Heisenberg picture

8. Mixed states and the density matrix

9. Perfect and imperfect measurements

10. Composite systems and entanglement

11. Quantum teleportation

12. Open quantum systems and completely positive maps

13. Orbital angular momentum

14. Spin angular momentum

15. Total angular momentum

16. Identical particles

17. Bose-Einstein and Fermi-Dirac statistics

18. Spin waves in solids

19. An atom in a cavity

20. Quantum field theory

21. Revision lecture (week before the exam)

PHY472: Advanced Quantum Mechanics

Contents

1 Linear Vector Spaces and Hilbert Space 81.1 Linear vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2 Operators in Hilbert space . . . . . . . . . . . . . . . . . . . . . . 101.3 Hermitian and unitary operators . . . . . . . . . . . . . . . . . . 141.4 Projection operators and tensor products . . . . . . . . . . . . . 161.5 The trace and determinant of an operator . . . . . . . . . . . . . 19

2 The Postulates of Quantum Mechanics 24

3 Schrödinger and Heisenberg Pictures 34

4 Mixed States and the Density Operator 414.1 Mixed states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.2 Decoherence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.3 Imperfect measurements . . . . . . . . . . . . . . . . . . . . . . . 454.4 Generalised observables . . . . . . . . . . . . . . . . . . . . . . . 50

5 Composite Systems and Entanglement 535.1 Composite systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 535.2 Entanglement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 545.3 Quantum teleportation . . . . . . . . . . . . . . . . . . . . . . . . 60

6 Evolution of Open Quantum Systems 646.1 The Lindblad equation . . . . . . . . . . . . . . . . . . . . . . . . 646.2 Positive and completely positive maps . . . . . . . . . . . . . . . 666.3 The Choi-Jamiołkowski isomorphism . . . . . . . . . . . . . . . . 67

7 Orbital Angular Momentum and Spin 717.1 Orbital angular momentum . . . . . . . . . . . . . . . . . . . . . 717.2 Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 757.3 Total angular momentum . . . . . . . . . . . . . . . . . . . . . . . 797.4 Composite systems with angular momentum . . . . . . . . . . . 80

5

8 Identical Particles 858.1 Symmetric and anti-symmetric states . . . . . . . . . . . . . . . 858.2 Creation and annihilation operators . . . . . . . . . . . . . . . . 878.3 The spin-statistics theorem . . . . . . . . . . . . . . . . . . . . . . 948.4 Bose-Einstein and Fermi-Dirac statistics . . . . . . . . . . . . . 95

9 Many-Body Problems in Quantum Mechanics 999.1 Interacting electrons in atomic shells . . . . . . . . . . . . . . . . 999.2 Spin waves in solids . . . . . . . . . . . . . . . . . . . . . . . . . . 1039.3 An atom in a cavity . . . . . . . . . . . . . . . . . . . . . . . . . . 1069.4 Outlook: quantum field theory . . . . . . . . . . . . . . . . . . . . 110

Suggested Further Reading

1. Quantum Processes, Systems, and Information, by Schumacherand Westmoreland, Cambridge University Press (2010). This is anexcellent book, and should be your first choice for additional material.It has everything up to relativistic quantum mechanics.

2. Quantum Information and Quantum Computation, by Nielsenand Chuang, Cambridge University Press (2000). This is the currentstandard work on quantum information theory. It has a comprehen-sive introduction to quantum mechanics along the lines treated here,but in more depth. The book is from 2000, which means that severalimportant recent topics are not covered.

3. Introductory Quantum Optics, by Gerry and Knight, CambridgeUniversity Press (2005). This is a very accessible introduction to thequantum theory of light.

4. Quantum Field Theory, by Lewis Ryder, Cambridge University Press(1996). This is a quite advanced introduction to relativistic quantumfield theory.

6

I would like to thank the students who have used these lecture notes inprevious years and spotted typos or errors (notably Tom Bullock). Theirefforts have greatly improved the readability of these notes.

7

1 Linear Vector Spaces and Hilbert Space

The modern version of quantum mechanics was formulated in 1932 by Johnvon Neumann in his famous book Mathematical Foundations of QuantumMechanics, and it unifies Schrödingers wave theory with the matrix mechan-ics of Heisenberg, Born, and Jordan. The theory is framed in terms of linearvector spaces, so the first couple of lectures we have to remind ourselves ofthe relevant mathematics.

1.1 Linear vector spaces

Consider a set of vectors, denoted by∣∣ψ⟩

,∣∣φ⟩

, etc., and the complex numbersa, b, c, etc. A linear vector space V is a mathematical structure of vectorsand numbers that obeys the following rules:

1.∣∣ψ⟩+ ∣∣φ⟩= ∣∣φ⟩+ ∣∣ψ⟩

(commutativity),

2.∣∣ψ⟩+ (

∣∣φ⟩+ ∣∣χ⟩)= (

∣∣ψ⟩+ ∣∣φ⟩)+

∣∣χ⟩(associativity),

3. a(∣∣ψ⟩+ ∣∣φ⟩

)= a∣∣ψ⟩+a

∣∣φ⟩(linearity),

4. (a+b)∣∣ψ⟩= a

∣∣ψ⟩+b∣∣ψ⟩

(linearity),

5. a(b∣∣φ⟩

)= (ab)∣∣φ⟩

.

There is also a null vector 0 such that∣∣ψ⟩+0 =

∣∣ψ⟩, and for every

∣∣ψ⟩there

is a vector∣∣φ⟩

such that∣∣ψ⟩+ ∣∣φ⟩= 0.

For each vector∣∣φ⟩

there is a dual vector⟨φ

∣∣, and the set of dual vectorsalso form a linear vector space V ∗. There is an inner product between vec-tors from V and V ∗ denoted by ⟨ψ|φ⟩. The inner product has the followingproperties:

1. ⟨ψ|φ⟩ = ⟨φ|ψ⟩∗,

2. ⟨ψ|ψ⟩ ≥ 0,

3. ⟨ψ|ψ⟩ = 0 ⇔∣∣ψ⟩= 0,

4.∣∣ψ⟩= c1

∣∣ψ1⟩+ c2

∣∣ψ2⟩ ⇒ ⟨φ|ψ⟩ = c1⟨φ|ψ1⟩+ c2⟨φ|ψ2⟩,

5. ∥φ∥≡√⟨φ|φ⟩ is the norm of∣∣φ⟩

.

8

If ∥φ ∥= 1, the vector∣∣φ⟩

is a unit vector. The set of unit vectors eiϕ ∣∣ψ⟩

with ϕ ∈ [0,2π) form a so-called ray in the linear vector space. A linear vectorspace that has a norm ∥ .∥ (there are many different ways we can define anorm) is called a Hilbert space. We will always assume that the linear vectorspaces are Hilbert spaces.

For linear vector spaces with an inner product we can derive the Cauchy-Schwarz inequality, also known as the Schwarz inequality:

|⟨φ|ψ⟩|2 ≤ ⟨ψ|ψ⟩⟨φ|φ⟩ . (1.1)

This is a very important relation, since it requires only the inner productstructure. Relations that are based on this inequality, such as the Heisen-berg uncertainty relation between observables, therefore have a very generalvalidity.

If two vectors have an inner product equal to zero, then these vectorsare called orthogonal. This is the definition of orthogonality. When thesevectors are also unit vectors, they are called orthonormal. A set of vectors∣∣φ1

⟩,∣∣φ2

⟩,. . .

∣∣φN⟩

are linearly independent if∑j

a j∣∣φ j

⟩= 0 (1.2)

implies that all a j = 0. The maximum number of linearly independent vec-tors in V is the dimension of V . Orthonormal vectors form a complete or-thonormal basis for V if any vector can be written as

∣∣ψ⟩= N∑k=1

ck∣∣φk

⟩, (1.3)

and ⟨φ j|φk⟩ = δ jk. We can take the inner product of∣∣ψ⟩

with any of the basisvectors

∣∣φ j⟩

to obtain

⟨φ j|ψ⟩ =N∑

k=1ck⟨φ j|φk⟩ =

N∑k=1

ckδ jk = c j . (1.4)

Substitute this back into the expansion of∣∣ψ⟩

, and we find

∣∣ψ⟩= N∑k=1

∣∣φk⟩⟨φk|ψ⟩ . (1.5)

9

Therefore∑

k∣∣φk

⟩⟨φk

∣∣ must act like the identity. In fact, this gives us animportant clue that operators of states must take the general form of sumsover objects like

∣∣φ⟩⟨χ∣∣.

1.2 Operators in Hilbert space

The objects∣∣ψ⟩

are vectors in a Hilbert space. We can imagine applyingrotations of the vectors, rescaling, permutations of vectors in a basis, and soon. These are described mathematically as operators, and we denote themby capital letters A, B, C, etc. In general we write

A∣∣φ⟩= ∣∣ψ⟩

, (1.6)

for some∣∣φ⟩

,∣∣ψ⟩ ∈ V . It is important to remember that operators act on all

the vectors in Hilbert space. Let ∣∣φ j

⟩ j be an orthonormal basis. We can

calculate the inner product between the vectors∣∣φ j

⟩and A

∣∣φk⟩:

⟨φ j|(A

∣∣φk⟩)= ⟨φ j|A|φk⟩ ≡ A jk . (1.7)

The two indices indicate that operators are matrices.As an example, consider two vectors, written as two-dimensional column

vectors ∣∣φ1⟩= (

10

),

∣∣φ2⟩= (

01

), (1.8)

and suppose that

A =(2 00 3

). (1.9)

We calculate

A11 =⟨φ1

∣∣ A∣∣φ1

⟩= (1,0

) ·(2 00 3

)(10

)= (

1,0) ·(2

0

)= 2. (1.10)

Similarly, we can calculate that A22 = 3, and A12 = A21 = 0 (check this). Wetherefore have that A

∣∣φ1⟩= 2

∣∣φ1⟩

and A∣∣φ2

⟩= 3∣∣φ2

⟩.

Complex numbers a have complex conjugates a∗ and vectors∣∣ψ⟩

havedual vectors

⟨φ

∣∣. Is there an equivalent for operators? The answer is yes,

10

and it is called the adjoint, or Hermitian conjugate, and is denoted by adagger †. The natural way to define it is according to the rule⟨

ψ∣∣ A

∣∣φ⟩∗ = ⟨φ

∣∣ A† ∣∣ψ⟩, (1.11)

for any∣∣φ⟩

and∣∣ψ⟩

. In matrix notation, and given an orthonormal basis∣∣φ j

⟩ j, this becomes⟨

φ j∣∣ A

∣∣φk⟩∗ = A∗

jk =⟨φk

∣∣ A† ∣∣φ j⟩= A†

k j . (1.12)

So the matrix representation of the adjoint A† is the transpose and the com-plex conjugate of the matrix A, as given by (A†) jk = A∗

k j. The adjoint has thefollowing properties:

1. (cA)† = c∗A†,

2. (AB)† = B† A†,

3. (∣∣φ⟩

)† = ⟨φ

∣∣.Note the order of the operators in 2: AB is generally not the same as BA.The difference between the two is called the commutator, denoted by

[A,B]= AB−BA . (1.13)

For example, we can choose

A =(0 11 0

)and B =

(1 00 −1

), (1.14)

which leads to

[A,B]=(0 11 0

)(1 00 −1

)−

(1 00 −1

)(0 11 0

)=

(0 −11 0

)−

(0 1−1 0

)=

(0 −22 0

)6= 0. (1.15)

Many, but not all, operators have an inverse. Let A∣∣φ⟩= ∣∣ψ⟩

and B∣∣ψ⟩= ∣∣φ⟩

.Then we have

BA∣∣φ⟩= ∣∣φ⟩

and AB∣∣ψ⟩= ∣∣ψ⟩

. (1.16)

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If Eq. (1.16) holds true for all∣∣φ⟩

and∣∣ψ⟩

, then B is the inverse of A, and wewrite B = A−1. An operator that has an inverse is called invertible. Anotherimportant property that an operator may possess is positivity. An operatoris positive if ⟨

φ∣∣ A

∣∣φ⟩≥ 0 for all∣∣φ⟩

. (1.17)

We also write this as A ≥ 0.From the matrix representation of operators you can easily see that the

operators themselves form a linear vector space:

1. A+B = B+ A,

2. A+ (B+C)= (A+B)+C,

3. a(A+B)= aA+aB,

4. (a+b)A = aA+bA,

5. a(bA)= (ab)A.

We can also define the operator norm ∥A∥ according to

∥A∥=√

Tr(A† A)≡√∑

i jA∗

i j A ji , (1.18)

which means that the linear vector space of operators is again a Hilbertspace. The symbol Tr(.) denotes the trace of an operator, and we will returnto this special operator property later in this section.

Every operator has a set of vectors for which

A∣∣a j

⟩= a j∣∣a j

⟩, with a j ∈C . (1.19)

This is called the eigenvalue equation (or eigenequation) for A, and the vec-tors

∣∣a j⟩

are the eigenvectors. The complex numbers a j are eigenvalues. Inthe basis of eigenvectors, the matrix representation of A becomes

A =

a1 0 · · · 00 a2 · · · 0...

.... . .

...0 0 · · · aN

(1.20)

12

When some of the a js are the same, we speak of degenerate eigenvalues.When there are n identical eigenvalues, we have n-fold degeneracy. Theeigenvectors corresponding to this eigenvalue then span an n-dimensionalsubspace of the vector space. We will return to subspaces shortly, when weintroduce projection operators.

For any orthonormal basis ∣∣φ j

⟩ j we have⟨

φ j∣∣ A

∣∣φk⟩= A jk , (1.21)

which can be written in the form

A =∑jk

A jk∣∣φ j

⟩⟨φk

∣∣ . (1.22)

For the special case where∣∣φ j

⟩= ∣∣a j⟩

this reduces to

A =∑

ja j

∣∣a j⟩⟨

a j∣∣ . (1.23)

This is the spectral decomposition of A. When all a j are equal, we havecomplete degeneracy over the full vector space, and the operator becomesproportional (up to a factor a j) to the identity I. Note that this is independentof the basis

∣∣a j⟩. As a consequence, for any orthonormal basis

∣∣φ j⟩ we

have

I=∑

j

∣∣φ j⟩⟨φ j

∣∣ . (1.24)

This is the completeness relation, and we will use this many times in ourcalculations.

Lemma: If two non-degenerate operators commute ([A,B] = 0), then theyhave a common set of eigenvectors.

Proof: Let A = ∑k ak |ak⟩⟨ak| and B = ∑

jk B jk∣∣a j

⟩⟨ak|. We can choose thiswithout loss of generality: we write both operators in the eigenbasis ofA. Furthermore, [A,B]= 0 implies that AB = BA.

AB =∑klm

akBlm |ak⟩⟨ak|al⟩⟨am| =∑lm

alBlm |al⟩⟨am|

BA =∑klm

akBlm |al⟩⟨am|ak⟩⟨ak| =∑lm

amBlm |al⟩⟨am| (1.25)

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Therefore

[A,B]=∑lm

(al −am)Blm |al⟩⟨am| = 0. (1.26)

If al 6= am for l 6= m, then Blm = 0, and Blm ∝ δlm. Therefore ∣∣a j

⟩ is

an eigenbasis for B. ä

The proof of the converse is left as an exercise. It turns out that this is alsotrue when A and/or B are degenerate.

1.3 Hermitian and unitary operators

Next, we will consider two special types of operators, namely Hermitian andunitary operators. An operator A is Hermitian if and only if A† = A.

Lemma: An operator is Hermitian if and only if it has real eigenvalues:A† = A ⇔ a j ∈R.

Proof: The eigenvalue equation of A implies that

A∣∣a j

⟩= a j∣∣a j

⟩ ⇒ ⟨a j

∣∣ A† = a∗j⟨a j

∣∣ , (1.27)

which means that⟨a j

∣∣ A∣∣a j

⟩ = a j and⟨a j

∣∣ A†∣∣a j

⟩ = a∗j . It is now

straightforward to show that A = A† implies a j = a∗j , or a j ∈ R. Con-

versely, a j ∈R implies a j = a∗j , and

⟨a j

∣∣ A∣∣a j

⟩= ⟨a j

∣∣ A† ∣∣a j⟩

. (1.28)

Let∣∣ψ⟩=∑

k ck |ak⟩. Then⟨ψ

∣∣ A∣∣ψ⟩=∑

j|c j|2

⟨a j

∣∣ A∣∣a j

⟩=∑j|c j|2

⟨a j

∣∣ A† ∣∣a j⟩=∑

j|c j|2

⟨a j

∣∣ A† ∣∣a j⟩

= ⟨ψ

∣∣ A† ∣∣ψ⟩(1.29)

for all∣∣ψ⟩

, and therefore A = A†. ä

14

Next, we construct the exponent of an operator A according to U = exp(icA).We have included the complex number c for completeness. At first sight, youmay wonder what it means to take the exponent of an operator. Recall, how-ever, that the exponent has a power expansion:

U = exp(icA)=∞∑

n=0

(ic)n

n!An . (1.30)

The nth power of an operator is straightforward: just multiply A n timeswith itself. The expression in Eq. (1.30) is then well defined, and the expo-nent is taken as an abbreviation of the power expansion. In general, we canconstruct any function of operators, as long as we can define the function interms of a power expansion:

f (A)=∞∑

n=0fn An . (1.31)

This can also be extended to functions of multiple operators, but now wehave to be very careful in the case where these operators do not commute.Familiar rules for combining normal functions no longer apply (see exercise4b).

We can construct the adjoint of the operator U according to

U† =( ∞∑

n=0

(ic)n

n!An

)†

=∞∑

n=0

(−ic∗)n

n!A†n = exp(−ic∗A†) . (1.32)

In the special case where A = A† and c is real, we calculate

UU† = exp(icA)exp(−ic∗A†)= exp(icA)exp(−icA)= exp[ic(A− A)]= I ,(1.33)

since A commutes with itself. Similarly, U†U = I. Therefore, U† =U−1, andan operator with this property is called unitary. Each unitary operator canbe generated by a Hermitian (self-adjoint) operator A and a real number c.A is called the generator of U . We often write U =UA(c). Unitary operatorsare basis transformations.

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1.4 Projection operators and tensor products

We can combine two linear vector spaces U and V into a new linear vectorspace W = U ⊕V . The symbol ⊕ is called the direct sum. The dimension ofW is the sum of the dimensions of U and V :

dimW = dimU +dimV . (1.34)

A vector in W can be written as

|Ψ⟩W =∣∣ψ⟩

U +∣∣φ⟩

V , (1.35)

where∣∣ψ⟩

U and∣∣φ⟩

V are typically not normalized (i.e., they are not unitvectors). The spaces U and V are so-called subspaces of W .

As an example, consider the three-dimensional Euclidean space spannedby the Cartesian axes x, y, and z. The xy-plane is a two-dimensional sub-space of the full space, and the z-axis is a one-dimensional subspace. Anythree-dimensional form can be projected onto the xy-plane by setting the zcomponent to zero. Similarly, we can project onto the z-axis by setting thex and y coordinates to zero. A projector is therefore associated with a sub-space. It acts on a vector in the full space, and forces all components to zero,except those of the subspace it projects onto.

The formal definition of a projector PU on U is given by

PU |Ψ⟩W =∣∣ψ⟩

U . (1.36)

This is equivalent to requiring that P2U

= PU , or PU is idempotent. One-dimensional projectors can be written as

P j =∣∣φ j

⟩⟨φ j

∣∣ . (1.37)

Two projectors P1 and P2 are orthogonal is P1P2 = 0. If P1P2 = 0, thenP1 +P2 is another projector:

(P1 +P2)2 = P21 +P1P2 +P2P1 +P2

2 = P21 +P2

2 = P1 +P2 . (1.38)

When P1 and P2 commute but are non-orthogonal (i.e., they overlap), thegeneral projector onto their combined subspace is

P1+2 = P1 +P2 −P1P2 . (1.39)

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(Prove this.) The orthocomplement of P is I−P, which is also a projector:

P(I−P)= P −P2 = P −P = 0 and (I−P)2 = I−2P +P2 = I−P . (1.40)

Another way to combine two vector spaces U and V is via the tensorproduct: W =U ⊗V , where the symbol ⊗ is called the direct product or tensorproduct. The dimension of the space W is then

dimW = dimU ·dimV . (1.41)

Let∣∣ψ⟩ ∈U and

∣∣φ⟩ ∈ V . Then∣∣ψ⟩⊗ ∣∣φ⟩ ∈W =U ⊗V . (1.42)

If∣∣ψ⟩=∑

j a j∣∣ψ j

⟩and

∣∣φ⟩=∑j b j

∣∣φ j⟩, then the tensor product of these vec-

tors can be written as∣∣ψ⟩⊗ ∣∣φ⟩=∑jk

a jbk∣∣ψ j

⟩⊗ ∣∣φk⟩=∑

jka jbk

∣∣ψ j⟩∣∣φk

⟩=∑jk

a jbk∣∣ψ j,φk

⟩, (1.43)

where we introduced convenient abbreviations for the tensor product nota-tion. The inner product of two vectors that are tensor products is

(⟨ψ1

∣∣⊗⟨φ1

∣∣)(∣∣ψ2⟩⊗ ∣∣φ2

⟩)= ⟨ψ1|ψ2⟩⟨φ1|φ2⟩ . (1.44)

Operators also obey the tensor product structure, with

(A⊗B)∣∣ψ⟩⊗ ∣∣φ⟩= (A

∣∣ψ⟩)⊗ (B

∣∣φ⟩) , (1.45)

and

(A⊗B)(C⊗D)∣∣ψ⟩⊗ ∣∣φ⟩= (AC

∣∣ψ⟩)⊗ (BD

∣∣φ⟩) . (1.46)

General rules for tensor products of operators are

1. A⊗0= 0⊗B = 0,

2. I⊗ I= I,

3. (A1 + A2)⊗B = A1 ⊗B+ A2 ⊗B,

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4. aA⊗bB = (ab)A⊗B,

5. (A⊗B)−1 = A−1 ⊗B−1,

6. (A⊗B)† = A† ⊗B†.

Note that the last rule preserves the order of the operators. In other words,operators always act on their own space. Often, it is understood implicitlywhich operator acts on which subspace, and we will write A ⊗ I = A andI⊗B = B. Alternatively, we can add subscripts to the operator, e.g., AU andBV .

As a practical example, consider two two-dimensional operators

A =(A11 A12A21 A22

)and B =

(B11 B12B21 B22

)(1.47)

with respect to some orthonormal bases |a1⟩ , |a2⟩ and |b1⟩ , |b2⟩ for A andB, respectively (not necessarily eigenbases). The question is now: what isthe matrix representation of A⊗B? Since the dimension of the new vectorspace is the product of the dimensions of the two vector spaces, we havedimW = 2 ·2= 4. A natural basis for A⊗B is then given by

∣∣a j,bk⟩ jk, with

j,k = 1,2, or

|a1⟩ |b1⟩ , |a1⟩ |b2⟩ , |a2⟩ |b1⟩ , |a2⟩ |b2⟩ . (1.48)

We can construct the matrix representation of A⊗B by applying this opera-tor to the basis vectors in Eq. (1.48), using

A∣∣a j

⟩= A1 j |a1⟩+ A2 j |a2⟩ and B |ak⟩ = B1k |b1⟩+B2k |b2⟩ , (1.49)

which leads to

A⊗B |a1,b1⟩ = (A11 |a1⟩+ A21 |a2⟩)(B11 |b1⟩+B21 |b2⟩)A⊗B |a1,b2⟩ = (A11 |a1⟩+ A21 |a2⟩)(B12 |b1⟩+B22 |b2⟩)A⊗B |a2,b1⟩ = (A12 |a1⟩+ A22 |a2⟩)(B11 |b1⟩+B21 |b2⟩)A⊗B |a2,b2⟩ = (A12 |a1⟩+ A22 |a2⟩)(B12 |b1⟩+B22 |b2⟩) (1.50)

18

Looking at the first line of Eq. (1.50), the basis vector |a1,b1⟩ gets mappedto all basis vectors:

A⊗B |a1,b1⟩ = A11B11 |a1,b1⟩+ A11B21 |a1,b2⟩+ A21B11 |a2,b1⟩+ A21B21 |a2,b2⟩ .(1.51)

Combining this into matrix form leads to

A⊗B =

A11B11 A11B12 A12B11 A12B12A11B21 A11B22 A12B21 A12B22A21B11 A21B12 A22B11 A22B12A21B21 A21B22 A22B21 A22B22

=(A11B A12BA21B A22B

). (1.52)

Recall that this is dependent on the basis that we have chosen. In particular,A⊗B may be diagonalized in some other basis.

1.5 The trace and determinant of an operator

There are two special functions of operators that play a key role in the the-ory of linear vector spaces. They are the trace and the determinant of anoperator, denoted by Tr(A) and det(A), respectively. While the trace and de-terminant are most conveniently evaluated in matrix representation, theyare independent of the chosen basis.

When we defined the norm of an operator, we introduced the trace. It isevaluated by adding the diagonal elements of the matrix representation ofthe operator:

Tr(A)=∑

j

⟨φ j

∣∣ A∣∣φ j

⟩, (1.53)

where ∣∣φ j

⟩ j is any orthonormal basis. This independence means that the

trace is an invariant property of the operator. Moreover, the trace has thefollowing important properties:

1. If A = A†, then Tr(A) is real,

2. Tr(aA)= aTr(A),

3. Tr(A+B)=Tr(A)+Tr(B),

19

4. Tr(AB)=Tr(BA) (the "cyclic property").

The first property follows immediately when we evaluate the trace in thediagonal basis, where it becomes a sum over real eigenvalues. The secondand third properties convey the linearity of the trace. The fourth property isextremely useful, and can be shown as follows:

Tr(AB)=∑

j

⟨φ j

∣∣ AB∣∣φ j

⟩=∑jk

⟨φ j

∣∣ A∣∣ψk

⟩⟨ψk

∣∣B∣∣φ j

⟩=

∑jk

⟨ψk

∣∣B∣∣φ j

⟩⟨φ j

∣∣ A∣∣ψk

⟩=∑k

⟨ψk

∣∣BA∣∣ψk

⟩=Tr(BA) . (1.54)

This derivation also demonstrates the usefulness of inserting a resolution ofthe identity in strategic places. In the cyclic property, the operators A andB may be products of two operators, which then leads to

Tr(ABC)=Tr(BCA)=Tr(CAB) . (1.55)

Any cyclic (even) permutation of operators under a trace gives rise to thesame value of the trace as the original operator ordering.

Finally, we construct the partial trace of an operator that lives on a ten-sor product space. Suppose that A ⊗B is an operator in the Hilbert spaceH1 ⊗H2. We can trace out Hilbert space H1, denoted by Tr1(.):

Tr1(A⊗B)=Tr(A)B , or equivalently Tr1(A1B2)=Tr(A1)B2 . (1.56)

Taking the partial trace has the effect of removing the entire Hilbert spaceH1 from the description. It reduces the total vector space. The partial tracealways carries an index, which determines which space is traced over.

The determinant of a 2×2 matrix is given by

det(A)= det(A11 A12A21 A22

)= A11 A22 − A12 A21 . (1.57)

The determinant of higher-dimensional matrices can be defined recursivelyas follows: The top-left element of an n×n matrix defines an (n−1)× (n−1)matrix by removing the top row and the left column. Similarly, any other

20

element in the left column defines an (n− 1)× (n− 1) matrix by removingthe left column and the row of the element we chose. The determinant ofthe n×n matrix is then given by the top-left element times the determinantof the remaining (n− 1)× (n− 1) matrix, minus the product of the secondelement down in the left column and the remaining (n−1)× (n−1) matrix,plus the third element times the remaining matrix, etc.

The determinant of the product of matrices is equal to the product of thedeterminants of the matrices:

det(AB)= det(A)det(B) . (1.58)

Moreover, if A is an invertible matrix, then we have

det(A−1)= det(A)−1 . (1.59)

This leads to an important relation between similar matrices A = X−1BX :

det(A)= det(X−1BX

)= det(X−1)

det(B)det(X )

= det(X )−1 det(B)det(X )= det(B) . (1.60)

In particular, this means that the determinant is independent of the basisin which the matrix is written, which means that it is an intrinsic propertyof the operator associated with that matrix.

Finally, here’s a fun relation between the trace and the determinant ofan operator:

det[exp(A)]= exp[Tr(A)] . (1.61)

Exercises

1. Vectors and matrices:

(a) Are the following three vectors linearly dependent or indepen-dent: a = (2,3,−1), b = (0,1,2), and c = (0,0,−5)?

(b) Consider the vectors∣∣ψ⟩= 3i

∣∣φ1⟩−7i

∣∣φ2⟩

and∣∣χ⟩= ∣∣φ1

⟩+2∣∣φ2

⟩,

with ∣∣φi

⟩ an orthonormal basis. Calculate the inner product

between∣∣ψ⟩

and∣∣χ⟩

, and show that they satisfy the Cauchy-Schwarz inequality.

21

(c) Consider the two matrices

A =0 i 2

0 1 01 0 0

and B =2 i 0

3 1 50 −i −2

. (1.62)

Calculate A−1B and BA−1. Are they equal?Hint: find the inverse of A in the diagonal basis.

(d) Calculate A⊗B and B⊗ A, where A =(0 11 0

)and B =

(1 00 −1

).

2. Operators:

(a) Which of these operators are Hermitian: A+ A†, i(A+ A†), i(A−A†), and A† A?

(b) Prove that a shared eigenbasis for two operators A and B impliesthat [A,B]= 0.

(c) Let U be a transformation matrix that maps one complete or-thonormal basis to another. Show that U is unitary.

(d) How many real parameters completely determine a d×d unitarymatrix?

3. Properties of the trace and the determinant:

(a) Calculate the trace and the determinant of the matrices A and Bin exercise 1c.

(b) Show that the expectation value of A can be written as Tr(∣∣ψ⟩⟨

ψ∣∣ A).

(c) Prove that the trace is independent of the basis.

4. Commutator identities.

(a) Let F(t)= eAteBt. Calculate dF/dt and use [eAt,B]= (eAtBe−At −B)eAt to simplify your result.

(b) Let G(t) = eAt+Bt+ f (t)H . Show by calculating dG/dt, and settingdF/dt = dG/dt at t = 1, that the following operator identity

eA eB = eA+B+ 12 [A,B] , (1.63)

22

holds if A and B both commute with [A,B]. Hint: use the Hada-mard lemma

eAtBe−At = B+ t1!

[A,B]+ t2

2![A, [A,B]]+ . . . (1.64)

(c) Show that the commutator of two Hermitian operators is anti-Hermitian (A† =−A).

(d) Prove the commutator analog of the Jacobi identity

[A, [B,C]]+ [B, [C, A]]+ [C, [A,B]]= 0. (1.65)

23

2 The Postulates of Quantum Mechanics

The entire structure of quantum mechanics (including its relativistic exten-sion) can be formulated in terms of states and operations in Hilbert space.We need rules that map the physical quantities such as states, observables,and measurements to the mathematical structure of vector spaces, vectorsand operators. There are several ways in which this can be done, and herewe summarize these rules in terms of five postulates.

Postulate 1: A physical system is described by a Hilbert space H , and thestate of the system is represented by a ray with norm 1 in H .

There are a number of important aspects to this postulate. First, the factthat states are rays, rather than vectors means that an overall phase eiϕ ofthe state does not have any physically observable consequences, and eiϕ ∣∣ψ⟩represents the same state as

∣∣ψ⟩. Second, the state contains all information

about the system. In particular, there are no hidden variables in this stan-dard formulation of quantum mechanics. Finally, the dimension of H maybe infinite, which is the case, for example, when H is the space of square-integrable functions.

As an example of this postulate, consider a two-level quantum system(a qubit). This system can be described by two orthonormal states |0⟩ and|1⟩. Due to linearity of Hilbert space, the superposition α |0⟩+β |1⟩ is againa state of the system if it has norm 1, or

(α∗ ⟨0|+β∗ ⟨1|)(α |0⟩+β |1⟩)= 1 or |α|2 +|β|2 = 1. (2.1)

This is called the superposition principle: any normalised superposition ofvalid quantum states is again a valid quantum state. It is a direct conse-quence of the linearity of the vector space, and as we shall see later, thisprinciple has some bizarre consequences that have been corroborated inmany experiments.

Postulate 2: Every physical observable A corresponds to a self-adjoint (Her-mitian1) operator A whose eigenvectors form a complete basis.

1In Hilbert spaces of infinite dimensionality, there are subtle differences between self-adjoint and Hermitian operators. We ignore these subtleties here, because we will be mostlydealing with finite-dimensional spaces.

24

We use a hat to distinguish between the observable and the operator, butusually this distinction is not necessary. In these notes, we will use hatsonly when there is a danger of confusion.

As an example, take the operator X :

X |0⟩ = |1⟩ and X |1⟩ = |0⟩ . (2.2)

This operator can be interpreted as a bit flip of a qubit. In matrix notationthe state vectors can be written as

|0⟩ =(10

)and |1⟩ =

(01

), (2.3)

which means that X is written as

X =(0 11 0

)(2.4)

with eigenvalues ±1. The eigenstates of X are

|±⟩ = |0⟩± |1⟩p2

. (2.5)

These states form an orthonormal basis.

Postulate 3: The eigenvalues of A are the possible measurement outcomes,and the probability of finding the outcome a j in a measurement isgiven by the Born rule:

p(a j)= |⟨a j|ψ⟩|2 , (2.6)

where∣∣ψ⟩

is the state of the system, and∣∣a j

⟩is the eigenvector as-

sociated with the eigenvalue a j via A∣∣a j

⟩ = a j∣∣a j

⟩. If a j is m-fold

degenerate, then

p(a j)=m∑

l=1|⟨a(l)

j |ψ⟩|2 , (2.7)

where the∣∣∣a(l)

j

⟩span the m-fold degenerate subspace.

25

The expectation value of A with respect to the state of the system∣∣ψ⟩

isdenoted by ⟨A⟩, and evaluated as

⟨A⟩ = ⟨ψ

∣∣ A∣∣ψ⟩= ⟨

ψ∣∣(∑

ja j

∣∣a j⟩⟨

a j∣∣)∣∣ψ⟩=∑

jp(a j)a j . (2.8)

This is the weighted average of the measurement outcomes. The spread ofthe measurement outcomes (or the uncertainty) is given by the variance

(∆A)2 = ⟨(A−⟨A⟩)2⟩ = ⟨A2⟩−⟨A⟩2 . (2.9)

So far we mainly dealt with discrete systems on finite-dimensional Hilbertspaces. But what about continuous systems, such as a particle in a box, ora harmonic oscillator? We can still write the spectral decomposition of anoperator A but the sum must be replace by an integral:

A =∫

da fA(a) |a⟩⟨a| , (2.10)

where |a⟩ is an eigenstate of A. Typically, there are problems with the nor-malization of |a⟩, which is related to the impossibility of preparing a systemin exactly the state |a⟩. We will not explore these subtleties further in thiscourse, but you should be aware that they exist. The expectation value of Ais

⟨A⟩ = ⟨ψ

∣∣ A∣∣ψ⟩= ∫

da fA(a)⟨ψ|a⟩⟨a|ψ⟩ ≡∫

da fA(a)|ψ(a)|2 , (2.11)

where we defined the wave function ψ(a) = ⟨a|ψ⟩, and |ψ(a)|2 is properlyinterpreted as the probability density that you remember from second-yearquantum mechanics.

The probability of finding the eigenvalue of an operator A in the intervala and a+da given the state

∣∣ψ⟩is⟨

ψ∣∣ (|a⟩⟨a|da)

∣∣ψ⟩≡ dp(a) , (2.12)

since both sides must be infinitesimal. We therefore find that

dp(a)da

= |ψ(a)|2 . (2.13)

26

Postulate 4: The dynamics of quantum systems is governed by unitarytransformations.

We can write the state of a system at time t as∣∣ψ(t)

⟩, and at some time t0 < t

as∣∣ψ(t0)

⟩. The fourth postulate tells us that there is a unitary operator

U(t, t0) that transforms the state at time t0 to the state at time t:∣∣ψ(t)⟩=U(t, t0)

∣∣ψ(t0)⟩

. (2.14)

Since the evolution from time t to t is denoted by U(t, t) and must be equalto the identity, we deduce that U depends only on time differences: U(t, t0)=U(t− t0), and U(0)= I.

As an example, let U(t) be generated by a Hermitian operator A accord-ing to

U(t)= exp(− iħAt

). (2.15)

The argument of the exponential must be dimensionless, so A must be pro-portional to ħ times an angular frequency (in other words, an energy). Sup-pose that

∣∣ψ(t)⟩

is the state of a qubit, and that A = ħωX . If∣∣ψ(0)

⟩= |0⟩ wewant to calculate the state of the system at time t. We can write

∣∣ψ(t)⟩=U(t)

∣∣ψ(0)⟩= exp(−iωtX ) |0⟩ =

∞∑n=0

(−iωt)n

n!X n . (2.16)

Observe that X2 = I, so we can separate the power series into even and oddvalues of n:

∣∣ψ(t)⟩= ∞∑

n=0

(−iωt)2n

(2n)!|0⟩+

∞∑n=0

(−iωt)2n+1

(2n+1)!X |0⟩ = cos(ωt) |0⟩− isin(ωt) |1⟩ .

(2.17)

In other words, the state oscillates between |0⟩ and |1⟩.The fourth postulate also leads to the Schrödinger equation. Let’s take

the infinitesimal form of Eq. (2.14):∣∣ψ(t+dt)⟩=U(dt)

∣∣ψ(t)⟩

. (2.18)

27

We require that U(dt) is generated by some Hermitian operator H:

U(dt)= exp(− iħH dt

). (2.19)

H must have the dimensions of energy, so we identify it with the energy oper-ator, or the Hamiltonian. We can now take a Taylor expansion of

∣∣ψ(t+dt)⟩

to first order in dt: ∣∣ψ(t+dt)⟩= ∣∣ψ(t)

⟩+dtddt

∣∣ψ(t)⟩+ . . . , (2.20)

and we expand the unitary operator to first order in dt as well:

U(dt)= 1− iħHdt+ . . . (2.21)

We combine this into∣∣ψ(t)⟩+dt

ddt

∣∣ψ(t)⟩= (

1− iħHdt

)∣∣ψ(t)⟩

, (2.22)

which can be recast into the Schrödinger equation:

iħ ddt

∣∣ψ(t)⟩= H

∣∣ψ(t)⟩

. (2.23)

Therefore, the Schrödinger equation follows directly from the postulates!

Postulate 5: If a measurement of an observable A yields an eigenvalue a j,then immediately after the measurement, the system is in the eigen-state

∣∣a j⟩

corresponding to the eigenvalue.

This is the infamous projection postulate, so named because a measurement“projects” the system to the eigenstate corresponding to the measured value.This postulate has as observable consequence that a second measurementimmediately after the first will also find the outcome a j. Each measure-ment outcome a j corresponds to a projection operator P j on the subspacespanned by the eigenvector(s) belonging to a j. A (perfect) measurement canbe described by applying a projector to the state, and renormalize:

∣∣ψ⟩→ P j∣∣ψ⟩

∥P j∣∣ψ⟩∥ . (2.24)

28

This also works for degenerate eigenvalues.We have established earlier that the expectation value of A can be writ-

ten as a trace:

⟨A⟩ =Tr(∣∣ψ⟩⟨

ψ∣∣ A) . (2.25)

Now instead of the full operator A, we calculate the trace of P j =∣∣a j

⟩⟨a j

∣∣:⟨P j⟩ =Tr(

∣∣ψ⟩⟨ψ

∣∣P j)=Tr(∣∣ψ⟩⟨ψ|a j⟩

⟨a j

∣∣)= |⟨a j|ψ⟩|2 = p(a j) . (2.26)

So we can calculate the probability of a measurement outcome by taking theexpectation value of the projection operator that corresponds to the eigen-state of the measurement outcome. This is one of the basic calculations inquantum mechanics that you should be able to do.

The projection postulate is somewhat problematic for the interpretationof quantum mechanics, because it leads to the so-called measurement prob-lem: Why does a measurement induce a non-unitary evolution of the system?After all, the measurement apparatus can also be described quantum me-chanically2 and then the system plus the measurement apparatus evolvesunitarily. But then we must invoke a new device that measures the com-bined system and measurement apparatus. However, this in turn can bedescribed quantum mechanically, and so on.

On the other hand, we do see definite measurement outcomes when wedo experiments, so at some level the projection postulate is necessary, andsomewhere there must be a “collapse of the wave function”. Schrödingeralready struggled with this question, and came up with his famous thoughtexperiment about a cat in a box with a poison-filled vial attached to a Geigercounter monitoring a radioactive atom (see Fig. 1). When the atom decays,it will trigger the Geiger counter, which in turn causes the release of thepoison killing the cat. When we do not look inside the box (more precisely:when no information about the atom-counter-vial-cat system escapes fromthe box), the entire system is in a quantum superposition. However, whenwe open the box, we do find the cat either dead or alive. One solution of the

2This is something most people require from a fundamental theory: quantum mechan-ics should not just break down for macroscopic objects. Indeed, experimental evidence ofmacroscopic superpositions has been found in the form of “cat states”.

29

Figure 1: Schrödingers Cat.

problem seems to be that the quantum state represents our knowledge ofthe system, and that looking inside the box merely updates our informationabout the atom, counter, vial and the cat. So nothing “collapses” except ourown state of mind.

However, this cannot be the entire story, because quantum mechanicsclearly is not just about our opinions of cats and decaying atoms. In par-ticular, if we prepare an electron in a spin “up” state |↑⟩, then wheneverwe measure the spin along the z-direction we will find the measurementoutcome “up”, no matter what we think about electrons and quantum me-chanics. So there seems to be some physical property associated with theelectron that determines the measurement outcome and is described by thequantum state.

Various interpretations of quantum mechanics attempt to address these(and other) issues. The original interpretation of quantum mechanics wasmainly put forward by Niels Bohr, and is called the Copenhagen interpre-tation. Broadly speaking, it says that the quantum state is a convenientfiction, used to calculate the results of measurement outcomes, and thatthe system cannot be considered separate from the measurement appara-tus. Alternatively, there are interpretations of quantum mechanics, such asthe Ghirardi-Rimini-Weber interpretation, that do ascribe some kind of re-ality to the state of the system, in which case a physical mechanism for thecollapse of the wave function must be given. Many of these interpretationscan be classified as hidden variable theories, which postulate that there is a

30

deeper physical reality described by some “hidden variables” that we mustaverage over. This in turn explains the probabilistic nature of quantum me-chanics. The problem with such theories is that these hidden variables mustbe quite weird: they can change instantly depending on events lightyearsaway3, thus violating Einstein’s theory of special relativity. Many physicistsdo not like this aspect of hidden variable theories.

Alternatively, quantum mechanics can be interpreted in terms of “manyworlds”: the Many Worlds interpretation states that there is one state vectorfor the entire universe, and that each measurement splits the universe intodifferent branches corresponding to the different measurement outcomes. Itis attractive since it seems to be a philosophically consistent interpretation,and while it has been acquiring a growing number of supporters over re-cent years4, a lot of physicists have a deep aversion to the idea of paralleluniverses.

Finally, there is the epistemic interpretation, which is very similar tothe Copenhagen interpretation in that it treats the quantum state to a largeextent as a measure of our knowledge of the quantum system (and the mea-surement apparatus). At the same time, it denies a deeper underlying real-ity (i.e., no hidden variables). The attractive feature of this interpretationis that it requires a minimal amount of fuss, and fits naturally with currentresearch in quantum information theory. The downside is that you have toabandon simple scientific realism that allows you to talk about the proper-ties of electrons and photons, and many physicists are not prepared to dothat.

As you can see, quantum mechanics forces us to abandon some deeplyheld (classical) convictions about Nature. Depending on your preference,you may be drawn to one or other interpretation. It is currently not knowwhich interpretation is the correct one.

3. . . even though the averaging over the hidden variables means you can never signalfaster than light.

4There seems to be some evidence that the Many Worlds interpretation fits well with thelatest cosmological models based on string theory.

31

Exercises

1. Calculate the eigenvalues and the eigenstates of the bit flip operatorX , and show that the eigenstates form an orthonormal basis. Calcu-late the expectation value of X for

∣∣ψ⟩= 1/p

3 |0⟩+ ip

2/3 |1⟩.

2. Show that the variance of A vanishes when∣∣ψ⟩

is an eigenstate of A.

3. Prove that an operator is Hermitian if and only if it has real eigenval-ues.

4. Show that a qubit in an unknown state∣∣ψ⟩

cannot be copied. This isthe no-cloning theorem. Hint: start with a state

∣∣ψ⟩ |i⟩ for some initialstate |i⟩, and require that for

∣∣ψ⟩= |0⟩ and∣∣ψ⟩= |1⟩ the cloning proce-

dure is a unitary transformation |0⟩ |i⟩→ |0⟩ |0⟩ and |1⟩ |i⟩→ |1⟩ |1⟩.

5. The uncertainty principle.

(a) Use the Cauchy-Schwarz inequality to derive the following rela-tion between non-commuting observables A and B:

(∆A)2(∆B)2 ≥ 14|⟨[A,B]⟩|2 . (2.27)

Hint: define | f ⟩ = (A −⟨A⟩)∣∣ψ⟩

and |g⟩ = i(B−⟨B⟩)∣∣ψ⟩

, and usethat |⟨ f |g⟩| ≥ 1

2 |⟨ f |g⟩+⟨g| f ⟩|.(b) Show that this reduces to Heisenberg’s uncertainty relation when

A and B are canonically conjugate observables, for example posi-tion and momentum.

(c) Does this method work for deriving the uncertainty principle be-tween energy and time?

6. Consider the Hamiltonian H and the state∣∣ψ⟩

given by

H = E

0 i 0−i 0 00 0 −1

and∣∣ψ⟩= 1p

5

1− i1− i

1

. (2.28)

where E is a constant with dimensions of energy. Calculate the energyeigenvalues and the expectation value of the Hamiltonian.

32

7. Show that the momentum and the total energy can be measured si-multaneously only when the potential is constant everywhere. Whatdoes a constant potential mean in terms of the dynamics of a particle?

33

3 Schrödinger and Heisenberg Pictures

So far we have assumed that the quantum states∣∣ψ(t)

⟩describing the sys-

tem carry the time dependence. However, this is not the only way to keeptrack of the time evolution. Since all physically observed quantities are ex-pectation values, we can write

⟨A⟩ =Tr[∣∣ψ(t)

⟩⟨ψ(t)

∣∣ A]=Tr

[U(t)

∣∣ψ(0)⟩⟨ψ(0)

∣∣U†(t)A]

=Tr[∣∣ψ(0)

⟩⟨ψ(0)

∣∣U†(t)AU(t)]

(cyclic property)

≡Tr[∣∣ψ(0)

⟩⟨ψ(0)

∣∣ A(t)], (3.1)

where we defined the time-varying operator A(t) =U†(t)AU(t). Clearly, wecan keep track of the time evolution in the operators!

• Schrödinger picture: Keep track of the time evolution in the states,

• Heisenberg picture: Keep track of the time evolution in the operators.

We can label the states and operators “S” and “H” depending on the picture.For example, ∣∣ψH

⟩= ∣∣ψS(0)⟩

and AH(t)=U†(t)ASU(t) . (3.2)

The time evolution for states is given by the Schrödinger equation, sowe want a corresponding “Heisenberg equation” for the operators. First, weobserve that

U(t)= exp(− iħHt

), (3.3)

such that

ddt

U(t)=− iħHU(t) . (3.4)

Next, we calculate the time derivative of ⟨A⟩:ddt

Tr[∣∣ψS(t)

⟩⟨ψS(t)

∣∣ AS]= d

dt⟨ψS(t)

∣∣ AS∣∣ψS(t)

⟩= ddt

⟨ψH

∣∣ AH(t)∣∣ψH

⟩.

(3.5)

34

The last equation follows from Eq. (3.1). We can now calculate

ddt

⟨ψS(t)

∣∣ AS∣∣ψS(t)

⟩= ddt

⟨ψS(0)

∣∣U†(t)ASU(t)∣∣ψS(0)

⟩= ⟨

ψS(0)∣∣[U†(t)ASU(t)+U†(t)ASU(t)+U†(t)ASU(t)

]∣∣ψS(0)⟩

= ⟨ψH

∣∣[ iħHAH(t)− i

ħAH(t)H+ ∂AH(t)∂t

]∣∣ψH⟩

=− iħ

⟨ψH

∣∣ [AH(t),H]∣∣ψH

⟩+⟨ψH

∣∣ ∂AH(t)∂t

∣∣ψH⟩

= ⟨ψH

∣∣ dAH(t)dt

∣∣ψH⟩

. (3.6)

Since this must be true for all∣∣ψH

⟩, this is an operator identity:

dAH(t)dt

=− iħ [AH(t),H]+ ∂AH(t)

∂t. (3.7)

This is the Heisenberg equation. Note the difference between the “straightd” and the “curly ∂” in the time derivative and the partial time derivative,respectively. The partial derivative deals only with the explicit time depen-dence of the operator. In many cases (such as position and momentum) thisis zero.

We have seen that both the Schrödinger and the Heisenberg equationfollows from Von Neumann’s Hilbert space formalism of quantum mechan-ics. Consequently, we have proved that this formalism properly unifies bothSchrödingers wave mechanics, and Heisenberg, Born, and Jordans matrixmechanics.

As an example, consider a qubit with time evolution determined by the

Hamiltonian H = 12ħωZ, with Z =

(1 00 −1

). This may be a spin in a magnetic

field, for example, such that ω = −eB/mc. We want to calculate the timeevolution of the operator XH(t). Since we work in the Heisenberg picturealone, we will omit the subscript H. First, we evaluate the commutator inthe Heisenberg equation

iħ12

dXdt

= 12

[X ,H]=−iħω2

Y , (3.8)

35

where we defined Y =(0 −ii 0

). So now we must know the time evolution of

Y as well:

iħ12

dYdt

= 12

[Y ,H]= iħω2

X . (3.9)

These are two coupled linear equations, which are relatively easy to solve:

X =−ωY and Y =ωX and Z = 0. (3.10)

We can define two new operators S± = X ± iY , and obtain

S± =−ωY ± iωX =±iωS± . (3.11)

Solving these two equations yields S±(t)= S±(0) e±iωt, and this leads to

X (t)= S+(t)+S−(t)2

= S+(0) eiωt +S−(0) e−iωt

2

= 12

[X (0) eiωt + iY (0) eiωt + X (0) e−iωt − iY (0) e−iωt

]= X (0)cos(ωt)−Y (0)sin(ωt) . (3.12)

You are asked to show that Y (t)=Y (0)cos(ωt)+ X (0)sin(ωt) in exercise 3.

We now take∣∣ψH

⟩= |0⟩ and X (0) =(0 11 0

), Y (0) =

(0 −ii 0

). The expecta-

tion value of X (t) is then readily calculated to be

⟨0|X (t) |0⟩ = cos(ωt)⟨0|X (0) |0⟩−sin(ωt)⟨0|Y (0) |0⟩ = 0. (3.13)

Alternatively, when∣∣ψH

⟩= |±⟩, we find

⟨+|X (t) |+⟩ = cos(ωt) and ⟨+|Y (t) |+⟩ = sin(ωt) . (3.14)

36

This is a circular motion in time:

a

b

|+!|"!X (t)

!t

Figure 1: Fidelity between the logical zero and one states received by Rob as

a function of the squeezing parameter r for single and dual rail qubits.

1

The eigenstate of X (π/2) is point a, and the eigenstate of X (−π/2) is point b.Furthermore, X (±π/2)=∓Y (0), and the states at point a and b are thereforethe eigenstates of Y :∣∣ψa

⟩= |0⟩− i |1⟩p2

and∣∣ψb

⟩= |0⟩+ i |1⟩p2

. (3.15)

A natural question to ask is where the states |0⟩ and |1⟩ fit in this picture.These are the eigenstates of the operator Z, which we used to generate theunitary time evolution. Clearly the states on the circle never become either|0⟩ or |1⟩, so we need to add another dimension:

a

b

|+!|"!

|0!

|1!

Figure 1: Fidelity between the logical zero and one states received by Rob as

a function of the squeezing parameter r for single and dual rail qubits.

1

This is called the Bloch sphere, and operators are represented by straightlines through the origin. The axis of rotation for the straight lines thatrotate with time is determined by the Hamiltonian. In the above case the

37

Hamiltonian was proportional to Z, which means that the straight linesrotate around the axis through the eigensates of Z, which are |0⟩ and |1⟩.

Exercises

1. Show that for the Hamiltonian HS = HH .

2. The harmonic oscillator has the energy eigenvalue equation H |n⟩ =ħω(n+ 1

2 ) |n⟩.

(a) The classical solution of the harmonic oscillator is given by

|α⟩ = e−12 |α|2

∞∑n=0

αnp

n!|n⟩ , (3.16)

in the limit of |α| À 1. Show that |α⟩ is a properly normalizedstate for any α ∈C.

(b) Calculate the time-evolved state |α(t)⟩.(c) We introduce the ladder operators a |n⟩ = p

n |n−1⟩ and a† |n⟩ =pn+1 |n+1⟩. Show that the number operator defined by n |n⟩ =

n |n⟩ can be written as n = a†a.

(d) Write the coherent state |α⟩ as a superposition of ladder opera-tors acting on the ground state |0⟩.

(e) Note that the ground state is time-independent (U(t) |0⟩ = |0⟩).Calculate the time evolution of the ladder operators.

(f) Calculate the position q = (a+ a†)/2 and momentum p = −i(a−a†)/2 of the harmonic oscillator in the Heisenberg picture. Canyou identify the classical harmonic motion?

3. Let A be an operator given by A = a0I+ axX + ayY + azZ. Calculatethe matrix A(t) given the Hamiltonian H = 1

2ħωZ, and show that A isHermitian when the aµ are real.

4. The interaction picture.

(a) Let the Hamiltonian of a system be given by H = H0 +V , withH0 = p2/2m. Using

∣∣ψ(t)⟩

I =U†0(t)

∣∣ψ(t)⟩

S with U0(t)= exp(−iH0t/ħ),

38

calculate the time dependence of an operator in the interactionpicture AI (t).

(b) Defining HI (t)=U†0(t)VU0(t), show that

iħ ddt

∣∣ψ(t)⟩

I = HI (t)∣∣ψ(t)

⟩I . (3.17)

Is HI identical to HH and HS?

5. The time operator in quantum mechanics.

(a) Let H∣∣ψ⟩ = E

∣∣ψ⟩, and assume the existence of a time operator

conjugate to H, i.e., [H,T]= iħ. Show that

H ei$T ∣∣ψ⟩= (E−ħ$)ei$T ∣∣ψ⟩. (3.18)

(b) Given that $ ∈R, calculate the spectrum of H.

(c) The energy of a system must be bounded from below in order toavoid infinite decay to ever lower energy states. What does thismean for T?

6. Consider a three-level atom with two (degenerate) low-lying states |0⟩and |1⟩ with zero energy, and a high level |e⟩ (the “excited” state) withenergy ħω. The low levels are coupled to the excited level by opticalfields Ω0 cosω0t and Ω1 cosω1t, respectively.

(a) Give the (time-dependent) Hamiltonian H for the system.

(b) The time dependence in H is difficult to deal with, so we musttransform to the rotating frame via some unitary transformationU(t). Show that

H′ =U(t)HU†(t)− iħUdU†

dt.

You can use the Schrödinger equation with∣∣ψ⟩=U†

∣∣ψ′⟩.

(c) Calculate H′ if U(t) is given by

U(t)=1 0 0

0 e−i(ω0−ω1)t 00 0 e−iω0 t

.

39

Why can we ignore the remaining time dependence in H′? Thisis called the Rotating Wave Approximation.

(d) Calculate the λ= 0 eigenstate of H′ in the case where ω0 =ω1.

(e) Design a way to bring the atom from the state |0⟩ to |1⟩ withoutever populating the state |e⟩.

40

4 Mixed States and the Density Operator

So far we have considered states as vectors in Hilbert space that, accordingto the first postulate, contain all the information about the system. In real-ity, however, we very rarely have complete information about a system. Forexample, the system may have interacted with its environment, which in-troduces some uncertainty in our knowledge of the state of the system. Thequestion is therefore how we describe systems given incomplete information.Much of contemporary research in quantum mechanics is about gaining fullcontrol over the quantum system (meaning to minimize the interaction withthe environment). This includes the field of quantum information and com-putation. The concept of incomplete information is therefore central to mod-ern quantum mechanics.

4.1 Mixed states

First, we recall some properties of the trace:

• Tr(aA)= aTr(A),

• Tr(A+B)=Tr(A)+Tr(B).

Also remember that we can write the expectation value of A as

⟨A⟩ =Tr(∣∣ψ⟩⟨

ψ∣∣ A) , (4.1)

where∣∣ψ⟩

is the state of the system. It tells us everything there is to knowabout the system. But what if we don’t know everything?

As an example, consider that Alice prepares a qubit in the state |0⟩ or inthe state |+⟩ = (|0⟩+ |1⟩)/

p2 depending on the outcome of a balanced (50:50)

coin toss. How does Bob describe the state before any measurement? First,we cannot say that the state is 1

2 |0⟩+ 12 |+⟩, because this is not normalized!

The key to the solution is to observe that the expectation values mustbehave correctly. The expectation value ⟨A⟩ is the average of the eigenvaluesof A for a given state. If the state is itself a statistical mixture (as in theexample above), then the expectation values must also be averaged. So for

41

the example, we require that for any A

⟨A⟩ = 12⟨A⟩0 +

12⟨A⟩+ = 1

2Tr(|0⟩⟨0|A)+ 1

2Tr(|+⟩⟨+|A)

=Tr[(

12|0⟩⟨0|+ 1

2|+⟩⟨+|

)A

]≡Tr(ρA) , (4.2)

where we defined

ρ = 12|0⟩⟨0|+ 1

2|+⟩⟨+| . (4.3)

The statistical mixture is therefore properly described by an operator, ratherthan a simple vector. We can generalize this as

ρ =∑k

pk∣∣ψk

⟩⟨ψk

∣∣ , (4.4)

where the pk are probabilities that sum up to one (∑

k pk = 1) and the∣∣ψk

⟩are normalized states (not necessarily complete or orthogonal). Since ρ actsas a weight, or a density, in the expectation value, we call it the densityoperator. We can diagonalize ρ to find the spectral decomposition

ρ =∑

jλ j

∣∣λ j⟩⟨λ j

∣∣ , (4.5)

where ∣∣λ j

⟩ j forms a complete orthonormal basis, 0 ≤ λ j ≤ 1, and

∑jλ j = 1.

We can also show that ρ is a positive operator:⟨ψ

∣∣ρ ∣∣ψ⟩=∑jk

c∗j ck⟨λ j

∣∣ρ |λk⟩ =∑jk

c∗j ck⟨λ j

∣∣∑lλl |λl⟩⟨λl |λk⟩

=∑jkl

c∗j ckλl⟨λ j|λl⟩⟨λl |λk⟩ =∑jkl

c∗j ckλlδ jlδlk

=∑lλl |cl |2

≥ 0. (4.6)

In general, an operator ρ is a valid density operator if and only if it has thefollowing three properties:

42

1. ρ† = ρ,

2. Tr(ρ)= 1,

3. ρ ≥ 0.

The density operator is a generalization of the state of a quantum systemwhen we have incomplete information. In the special case where one of thep j = 1 and the others are zero, the density operator becomes the projector∣∣ψ j

⟩⟨ψ j

∣∣. In other words, it is completely determined by the state vector∣∣ψ j⟩. We call these pure states. The statistical mixture of pure states giving

rise to the density operator is called a mixed state.The unitary evolution of the density operator can be derived directly

from the Schrödinger equation iħ∂t∣∣ψ⟩= H

∣∣ψ⟩:

iħdρdt

= iħ ddt

∑j

p j∣∣ψ j

⟩⟨ψ j

∣∣= iħ

∑j

dp j

dt∣∣ψ j

⟩⟨ψ j

∣∣+ p j

[(ddt

∣∣ψ j⟩)⟨

ψ j∣∣+ ∣∣ψ j

⟩(ddt

⟨ψ j

∣∣)]= iħ∂ρ

∂t+Hρ−ρH

= [H,ρ]+ iħ∂ρ∂t

. (4.7)

This agrees with the Heisenberg equation for operators, and it is sometimesknown as the Von Neumann equation. In most problems the probabilities p jhave no explicit time-dependence, and ∂tρ = 0.

4.2 Decoherence

The density operator allows us to consider the phenomenon of decoherence.Consider the pure state |+⟩. In matrix notation with respect to the basis|+⟩ , |−⟩, this can be written as

ρ =(1 00 0

). (4.8)

43

The trace is 1, and one of the eigenvalues is 1, as required for a pure state.We can also write the density operator in the basis |0⟩ , |1⟩:

ρ = |+⟩⟨+| = 1p2

(11

)× 1p

2

(1,1

)= 12

(1 11 1

). (4.9)

Notice how the outer product (as opposed to the inner product) of two vectorscreates a matrix representation of the corresponding projection operator.

Let the time evolution of |+⟩ be given by

|+⟩ = |0⟩+ |1⟩p2

→ |0⟩+ eiωt |1⟩p2

. (4.10)

The corresponding density operator becomes

ρ(t)= 12

(1 eiωt

e−iωt 1

). (4.11)

The “population” in the state |+⟩ is given by the expectation value

⟨+|ρ(t) |+⟩ = 12+ 1

2cos(ωt) . (4.12)

This oscillation is due to the off-diagonal elements of ρ(t), and it is called thecoherence of the system (see Fig. 2). The state is pure at any time t. In realphysical systems the coherence often decays exponentially at a rate γ, andthe density matrix can be written as

ρ(t)= 12

(1 eiωt−γt

e−iωt−γt 1

). (4.13)

The population in the state |+⟩ decays accordingly as

⟨+|ρ(t) |+⟩ = 12+ e−γt cos(ωt)

2. (4.14)

This is called decoherence of the system, and the value of γ depends on thephysical mechanism that leads to the decoherence.

The decoherence described above is just one particular type, and is calleddephasing. Another important decoherence mechanism is relaxation to the

44

0 5 10 15 20 25 30 350.0

0.2

0.4

0.6

0.8

1.0

time

Pop

ula

tion

Figure 1:

1

Figure 2: Population in the state |+⟩ with decoherence (solid curve) andwithout (dashed curve).

ground state. If the state |1⟩ has a larger energy than |0⟩ there may be pro-cesses such as spontaneous emission that drive the system to the groundstate. Combining these two decay processes, we can write the density oper-ator as

ρ(t)= 12

(2− e−γ1 t eiωt−γ2 t

e−iωt−γ2 t e−γ1 t

). (4.15)

The study of decoherence is currently one of the most important researchareas in quantum physics.

4.3 Imperfect measurements

Postulates 3 and 5 determine what happens when a quantum system issubjected to a measurement. In particular, these postulates concern idealmeasurements. In practice, however, we often have to deal with imperfectmeasurements that include noise. As a simple example, consider a pho-todetector: not every photon that hits the detector will result in a detector“click”, which tells us that there indeed was a photon. How can we describesituations like these?

45

First, let’s recap ideal measurements. We can ask the question whatwill be the outcome of a single measurement of the observable A. We knowfrom Postulate 3 that the outcome m must be an eigenvalue am of A. If thespectral decomposition of A is given by

A =∑m

am |m⟩⟨m| , (4.16)

then the probability of finding measurement outcome m is given by the Bornrule

p(m)= |⟨m|ψ⟩|2 =Tr(Pm∣∣ψ⟩⟨

ψ∣∣)≡ ⟨Pm⟩ , (4.17)

where we introduced the operator Pm = |m⟩⟨m|. One key interpretation ofp(m) is as the expectation value of the operator Pm associated with mea-surement outcome m.

When a measurement does not destroy the system, the state of the sys-tem must be updated after the measurement to reflect the fact that anothermeasurement of A immediately following the first will yield outcome m withcertainty. The update rule is given by Eq. (2.24)

∣∣ψ⟩→ Pm∣∣ψ⟩√⟨

ψ∣∣P†

mPm∣∣ψ⟩ , (4.18)

where the square root in the denominator is included to ensure proper nor-malization of the state after the measurement. However, this form is notso easily generalized to measurements yielding incomplete information, soinstead we will write (using P2

m = Pm = P†m)

∣∣ψ⟩⟨ψ

∣∣→ Pm∣∣ψ⟩⟨

ψ∣∣P†

m

Tr(Pm∣∣ψ⟩⟨

ψ∣∣P†

m). (4.19)

From a physical perspective this notation is preferable, since the unobserv-able global phase of ⟨m|ψ⟩ no longer enters the description. Up to this point,both state preparation and measurement were assumed to be ideal. Howmust this formalism of ideal measurements be modified in order to take intoaccount measurements that yield only partial information about the system?

46

First, we must find the probability for measurement outcome m, and second,we have to formulate the update rule for the state after the measurement.

When we talk about imperfect measurements, we must have some wayof knowing (or suspecting) how the measurement apparatus fails. For ex-ample, we may suspect that a photon hitting a photodetector has only afinite probability of triggering a detector “click”. We therefore describe ourmeasurement device with a general probability distribution q j(m), whereq j(m) is the probability that the measurement outcome m in the detector istriggered by a system in the state

∣∣ψ j⟩. The probabilities q j(m) can be found

by modelling the physical aspects of the measurement apparatus. The accu-racy of this model can then be determined by experiment. The probabilityof measurement outcome m for the ideal case is given in Eq. (4.17) by ⟨Pm⟩.When the detector is not ideal, the probability of finding outcome m is givenby the weighted average over all possible expectation values ⟨P j⟩ that canlead to m:

p(m)=∑

jq j(m)⟨P j⟩ =

∑j

q j(m)Tr(P j∣∣ψ⟩⟨

ψ∣∣)

=Tr

[∑j

q j(m)P j∣∣ψ⟩⟨

ψ∣∣]≡Tr(Em

∣∣ψ⟩⟨ψ

∣∣)= ⟨Em⟩ , (4.20)

where we defined the operator Em associated with outcome m as

Em =∑

jq j(m)P j . (4.21)

Each possible measurement outcome m has an associated operator Em, theexpectation value of which is the probability of obtaining m in the measure-ment. The set of Em is called a Positive Operator-Valued Measure (POVM).While the total number of ideal measurement outcomes, modelled by Pm,must be identical to the dimension of the Hilbert space (ignoring the tech-nical complication of degeneracy), this is no longer the case for the POVM

described by Em; there can be more or fewer measurement outcomes, de-pending on the physical details of the measurement apparatus. For exam-ple, the measurement of an electron spin in a Stern-Gerlach apparatus canhave outcomes “up”, “down”, or “failed measurement”. The first two are

47

determined by the position of the fluorescence spot on the screen, and thelast may be the situation where the electron fails to produce a spot on thescreen. Here, the number of possible measurement outcomes is greater thanthe number of eigenstates of the spin operator. Similarly, photodetectors inGeiger mode have only two possible outcomes, namely a “click” or “no click”depending on whether the detector registered photons or not. However, thenumber of eigenstates of the intensity operator (the photon number states)is infinite.

Similar to the density operator, the POVM elements Em have three keyproperties. First, the p(m) are probabilities and therefore real, so for allstates

∣∣ψ⟩we have

⟨Em⟩∗ = ⟨Em⟩ ⇐⇒ ⟨ψ

∣∣Em∣∣ψ⟩= ⟨

ψ∣∣E†

m∣∣ψ⟩

, (4.22)

and Em is therefore Hermitian. Second, since∑

m p(m)= 1 we have∑m

p(m)=∑m

⟨ψ

∣∣Em∣∣ψ⟩= ⟨

ψ∣∣∑

mEm

∣∣ψ⟩= 1, (4.23)

for all∣∣ψ⟩

, and therefore ∑m

Em = I , (4.24)

where I is the identity operator. Finally, since p(m) = ⟨ψ

∣∣Em∣∣ψ⟩ ≥ 0 for all∣∣ψ⟩

, the POVM element Em is a positive operator. Note the close analogy ofthe properties of the POVM and the density operator. In particular, just as inthe case of density operators, POVMs are defined by these three properties.

The second question about generalized measurements is how the mea-surement outcomes should be used to update the quantum state of the sys-tem. The rule for ideal von Neumann measurements is given in Eq. (4.19),which can be generalized immediately to density operators using the tech-niques presented above. This yields

ρ→ PmρP†m

Tr(PmρP†m)

. (4.25)

What if we have instead a measurement apparatus described by a POVM?Consistency with the Born rule in Eq. (4.20) requires that we again replace

48

Pm, associated with measurement outcome m, with a probability distribu-tion over all P j:

ρ→∑

jq j(m)

P jρP†j

Tr[∑

k qk(m)PkρP†k

] =∑

jq j(m)

P jρP†j

Tr(Emρ), (4.26)

where we adjusted the normalization factor to maintain Tr(ρ) = 1. We alsoused that the POVM element Em in Eq. (4.21) can be written as

Em =∑

jq j(m)P†

j P j . (4.27)

If we rescale the P j by a factor√

q j(m), we obtain the standard form of thePOVM

Em =∑

jA†

jm A jm , (4.28)

where A jm =√q j(m)P j are the so-called the Kraus operators. Consequently,

the update rule can be written as

ρ→∑

j A jmρA†jm

Tr(Emρ), (4.29)

which generally yields a mixed state (described by a density operator) afteran incomplete measurement.

Finally, we can consider the case of some nonunitary evolution withouta measurement (e.g., when the system interacts with its environment). Wecan model this purely formally by removing the index of the measurementoutcome m from the description (since there are no measurement outcomes).The most general evolution then takes the form

ρ→ ρ′ =∑k

AkρA†k , (4.30)

and the question is what form the Kraus operators Ak take. We will returnto this in section 6.

49

4.4 Generalised observables

Previously we constructed observables from projection operators associatedwith orthonormal basis states:

A =∑

ja jP j , (4.31)

where a j are the (real) eigenvalues of A and the P j =∣∣a j

⟩⟨a j

∣∣ are the pro-jectors onto the eigenvectors

∣∣a j⟩. However, just as with imperfect measure-

ments, we can construct “imperfect” observables, where the projectors arereplaced with POVMs. These are called generalised observables.

As an example, consider the observable whose outcomes are images in acamera. For simplicity we take it to be a black-and-white camera. The imagerecorded with the camera will be a two-dimensional array of intensities thatmake up the picture. If we label each pixel by the position x, and each pixelhas a number of possible intensity values ik ≥ 0, the image is given by

I(x)=∑k

ik pk(x) , (4.32)

where pk(x) is the probability that the pixel at position x records intensityik. In quantum mechanics, pk(x) is given by the Born rule

pk(x)=Tr[ρ Ek(x)

], (4.33)

with ρ the quantum state of the light impinging on the camera, and Ek(x)the measurement operator for outcome ik at pixel x. This operator takesinto account saturation effects, pixel bleeding, and other imperfections inimaging systems. We can then combine Eqs. (4.32) and (4.33) into

I(x)=∑k

ik Tr[ρ Ek(x)

]=Tr

[ρ

∑k

ik Ek(x)

]≡Tr

[ρ I(x)

]. (4.34)

You now see that the image I(x) is the expectation value of the image opera-tor I(x), which is a generalised observable:

I(x)=∑k

ik Ek(x) . (4.35)

It can be decomposed into projectors, but only at the cost of losing the phys-ical intensities of the pixels ik as the eigenvalues.

50

Exercises

1. The density matrix.

(a) Show that 12 |0⟩+ 1

2 |+⟩ is not a properly normalized state.(b) Show that Tr(ρ)= 1 with ρ given by Eq. (4.3), and then prove that

any density operator has unit trace and is Hermitian.(c) Show that density operators are convex, i.e., that ρ = w1ρ1+w2ρ2

with w1 +w2 = 1 (w1,w2 ≥ 0), and ρ1, ρ2 again density operators.(d) Calculate the expectation value of A using the two representa-

tions of ρ in terms of pi and the spectral decomposition. What isthe difference in the physical interpretation of p j and λ j?

2. Using the identity ⟨x|A|ψ⟩ = Aψ(x), and the resolution of the identity∫dx |x⟩⟨x| = I, calculate the expectation value for an operator A, given

a mixed state of wave functions.

3. Calculate P2 with P given by

P =(ab

)(a∗,b∗)= (|a|2 ab∗

a∗b |b|2)

and |a|2 +|b|2 = 1. (4.36)

4. Calculate the eigenvalues of the density matrix in Eq. (4.15), and showthat γ1 ≤ 2γ2. Hint: it is difficult to derive the inequality directly, soyou should try to demonstrate that γ1 ≥ 2γ2 leads to a contradiction.

5. A system with energy eigenstates |En⟩ is in thermal equilibrium witha heat bath at temperature T. The probability for the system to be instate En is proportional to e−En/kT .

(a) Write the Hamiltonian of the system in terms of En and |En⟩.(b) Give the normalized density operator ρ for the system as a func-

tion of the Hamiltonian.(c) We identify the normalization of ρ with the partition function Z .

Calculate the average energy directly and via

⟨E⟩ =−∂ lnZ

∂(kT). (4.37)

of the system.

51

(d) What is the entropy S = k lnZ if the system is a harmonic oscil-lator? Comment on the limit T → 0.

6. Lossy photodetectors.

(a) The state of a beam of light can be written in the photon numberbasis |n⟩ as

∣∣ψ⟩ = ∑n cn |n⟩. What are the possible measurement

outcomes for a perfect photon detector? Calculate the probabili-ties of the measurement outcome using projection operators.

(b) Suppose that the detector can only tell the difference between thepresence and absence of photons (a so-called “bucket detector”).How do we calculate the probability of finding the measurementoutcomes?

(c) Real bucket detectors have a finite efficiency η, which means thateach photon has a probability η of triggering the detector. Calcu-late the probabilities of the measurement outcomes.

(d) What other possible imperfections do realistic bucket detectorshave?

7. An electron with spin state∣∣ψ⟩ = α |↑⟩+β |↓⟩ and |α|2 +|β|2 = 1 is sent

through a Stern-Gerlach apparatus to measure the spin in the z direc-tion (i.e., the |↑⟩, |↓⟩ basis).

(a) What are the possible measurement outcomes? If the positionof the electron is measured by an induction loop rather than ascreen, what is the state of the electron immediately after themeasurement?

(b) Suppose that with probability p the induction loops fail to givea current signifying the presence of an electron. What are thethree possible measurement outcomes? Give the POVM.

(c) Calculate the probabilities of the measurement outcomes, andthe state of the electron imediately after the measurement (forall possible outcomes).

52

5 Composite Systems and Entanglement

5.1 Composite systems

Suppose we have two systems, described by Hilbert spaces H1 and H2, re-spectively. We can choose orthonormal bases for each system:

H1 :∣∣φ1

⟩,∣∣φ2

⟩, . . . ,

∣∣φN⟩

and H2 :∣∣ψ1

⟩,∣∣ψ2

⟩, . . . ,

∣∣ψM⟩

. (5.1)

The respective dimensions of H1 and H2 are N and M. We can constructN ×M basis states for the composite system via

∣∣φ j⟩

and∣∣ψk

⟩. This implies

that the total Hilbert space of the composite system can be spanned by thetensor product ∣∣φ j

⟩⊗ ∣∣ψk⟩

jk on H1+2 =H1 ⊗H2 . (5.2)

An arbitrary pure state on H1+2 can be written as

|Ψ⟩ =∑jk

c jk∣∣φ j

⟩⊗ ∣∣ψk⟩≡∑

jkc jk

∣∣φ j,ψk⟩

. (5.3)

For example, the system of two qubits can be written on the basis |0,0⟩ , |0,1⟩,|1,0⟩ , |1,1⟩. If system 1 is in state

∣∣φ⟩and system 2 is in state

∣∣ψ⟩, the par-

tial trace over system 2 yields

Tr2(∣∣φ,ψ

⟩⟨φ,ψ

∣∣)=Tr2(∣∣φ⟩⟨

φ∣∣⊗ ∣∣ψ⟩⟨

ψ∣∣)= ∣∣φ⟩⟨

φ∣∣Tr(

∣∣ψ⟩⟨ψ

∣∣)= ∣∣φ⟩⟨φ

∣∣ ,(5.4)

since the trace over any density operator is 1. We have now lost system 2from our description! Therefore, taking the partial trace without insertingany other operators is the mathematical version of forgetting about it. Thisis a very useful feature: you often do not want to deal with every possiblesystem you are interested in. For example, if system 1 is a qubit, and systemtwo is a very large environment the partial trace allows you to “trace out theenvironment”.

However, tracing out the environment will not always leave you with apure state as in Eq. (5.4). If the system has interacted with the environment,taking the partial trace generally leaves you with a mixed state. This is dueto entanglement between the system and its environment.

53

5.2 Entanglement

Consider the following experiment: Alice and Bob each blindly draw a mar-ble from a vase that contains one black and one white marble. Let’s call thestate of the write marble |0⟩ and the state of the black marble |1⟩. If wedescribe this classical experiment quantum mechanically (we can always dothis, because classical physics is contained in quantum physics), then thereare two possible states, |1,0⟩ and |0,1⟩. Since blind drawing is a statisticalprocedure, the state of the marbles held by Alice and Bob is the mixed state

ρ = 12|0,1⟩⟨0,1|+ 1

2|1,0⟩⟨1,0| . (5.5)

From Alice’s perspective, the state of her marble is obtained by tracing overBob’s marble:

ρA =TrB(ρ)= 12|0⟩⟨0|+ 1

2|1⟩⟨1| . (5.6)

This is what we expect: Alice has a 50:50 probability of finding “white” or“black” when she looks at her marble (i.e., when she measures the colour ofthe marble).

Next, consider what the state of Bob’s marble is when Alice finds a whitemarble. Just from the setup we know that Bob’s marble must be black,because there was only one white and one black marble in the vase. Let’ssee if we can reproduce this in our quantum mechanical description. Findinga white marble can be described mathematically by a projection operator|0⟩⟨0| (see Eq. (2.24)). We need to include this operator in the trace overAlice’s marble’s Hilbert space:

ρB = TrA(|0⟩A ⟨0|ρ)Tr(|0⟩A ⟨0|ρ)

= |1⟩⟨1| , (5.7)

which we set out to prove: if Alice finds that when she sees that her marble iswhite, she describes the state of Bob’s marble as black. Based on the setup ofthis experiment, Alice knows instantaneously what the state of Bob’s marbleis as soon as she looks at her own marble. There is nothing spooky aboutthis; it just shows that the marbles held by Alice and Bob are correlated.

54

Next, consider a second experiment: By some procedure, the details ofwhich are not important right now, Alice and Bob each hold a two-level sys-tem (a qubit) in the pure state

|Ψ⟩AB = |0,1⟩+ |1,0⟩p2

. (5.8)

Since |1,0⟩ and |0,1⟩ are valid quantum states, by virtue of the first postulateof quantum mechanics |Ψ⟩AB is also a valid quantum mechanical state. Itis not difficult to see that these systems are also correlated in the states |0⟩and |1⟩: When Alice finds the value “0”, Bob must find the value “1”, and viceversa. We can write this state as a density operator

ρ = 12

(|0,1⟩+ |1,0⟩)(⟨0,1|+⟨1,0|)

= 12

(|0,1⟩⟨0,1|+ |0,1⟩⟨1,0|+ |1,0⟩⟨0,1|+ |1,0⟩⟨1,0|) . (5.9)

Notice the two extra terms with respect to Eq. (5.5). If Alice now traces outBob’s system, she finds that the state of her marble is

ρA =TrB(ρ)= 12|0⟩⟨0|+ 1

2|1⟩⟨1| . (5.10)

In other words, even though the total system was in a pure state, the sub-system held by Alice (and Bob, check this) is mixed! We can try to put thetwo states back together:

ρA ⊗ρB =(

12|0⟩⟨0|+ 1

2|1⟩⟨1|

)⊗

(12|0⟩⟨0|+ 1

2|1⟩⟨1|

)= 1

4(|0,0⟩⟨0,0|+ |0,1⟩⟨0,1|+ |1,0⟩⟨1,0|+ |1,1⟩⟨1,1|) , (5.11)

but this is not the state we started out with! It is also a mixed state, in-stead of the pure state we started with. Since mixed states mean incompleteknowledge, there must be some information in the combined system thatdoes not reside in the subsystems alone! This is called entanglement.

Entanglement arises because states like (|0,1⟩+|1,0⟩)/p

2 cannot be writ-ten as a tensor product of two pure states

∣∣ψ⟩⊗ ∣∣φ⟩. These latter states are

55

called separable. In general a state is separable if and only if it can be writ-ten as

ρ =∑

jp j ρ

(A)j ⊗ρ(B)

j . (5.12)

Classical correlations such as the black and white marbles above fall intothe category of separable states.

So far, we have considered the quantum states in the basis |0⟩ , |1⟩.However, we can also describe the same system in the rotated basis |±⟩according to

|0⟩ = |+⟩+ |−⟩p2

and |1⟩ = |+⟩− |−⟩p2

. (5.13)

The entangled state |Ψ⟩AB can then be written as

|0,1⟩+ |1,0⟩p2

= |+,+⟩−|−,−⟩p2

, (5.14)

which means that we have again perfect correlations between the two sys-tems with respect to the states |+⟩ and |−⟩. Let’s do the same for the state ρin Eq. (5.5) for classically correlated marbles. After a bit of algebra, we findthat

ρ =14

(|++⟩⟨++|+ |+−⟩⟨+−|+ |−+⟩⟨−+|+ |−−⟩⟨−−|−|++⟩⟨−−|− |−−⟩⟨++|− |+−⟩⟨−+|− |−+⟩⟨+−|) . (5.15)

Now there are no correlations in the conjugate basis |±⟩, which you cancheck by calculating the conditional probabilities of Bob’s state given Alice’smeasurement outcomes. This is another key difference between classicallycorrelated states and entangled states. A good interpretation of entangle-ment is that entangled systems exhibit correlations that are stronger thanclassical correlations. We will shortly see how these stronger correlationscan be used in information processing.

We have seen that operators, just like states, can be combined into tensorproducts:

A⊗B∣∣φ⟩⊗ ∣∣ψ⟩= A

∣∣φ⟩⊗B∣∣ψ⟩

. (5.16)

56

And just like states, some operators cannot be written as A⊗B:

C =∑k

Ak ⊗Bk . (5.17)

This is the most general expression of an operator in the Hilbert space H1⊗H2. In Dirac notation this becomes

C =∑

jklmφ jklm

∣∣φ j⟩⟨φk

∣∣⊗ ∣∣φl⟩⟨φm

∣∣= ∑jklm

φ jklm∣∣φ j,φl

⟩⟨φk,φm

∣∣ . (5.18)

As an example, the Bell operator is diagonal on the Bell basis:

∣∣Φ±⟩= |0,0⟩± |1,1⟩p2

and∣∣Ψ±⟩= |0,1⟩± |1,0⟩p

2. (5.19)

The eigenvalues of the Bell operator are not important, as long as they arenot degenerate (why?). A measurement of the Bell operator projects onto aneigenstate of the operator, which is an entangled state. Consequently, wecannot implement such composite measurements by measuring each sub-system individually, because those individual measurements would projectonto pure states of the subsystems. And we have seen that the subsystemsof pure entangled states are mixed states.

A particularly useful technique when dealing with two systems is the so-called Schmidt decomposition. In general, we can write any pure state overtwo systems as a superposition of basis states:

|Ψ⟩ =dA∑j=1

dB∑k=1

c jk∣∣φ j

⟩A

∣∣ψk⟩

B , (5.20)

where dA and dB are the dimensions of the Hilbert spaces of system A andB, respectively, and we order the systems such that dB ≥ dA. It turns outthat we can always simplify this description and write |Ψ⟩ as a single sumover basis states. We state it as a theorem:

Theorem: Let |Ψ⟩ be a pure state of two systems, A and B with Hilbertspaces HA and HB of dimension dA and dB ≥ dA, respectively. There

57

exist orthonormal basis vectors∣∣a j

⟩A for system A and

∣∣b j⟩

B for sys-tem B such that

|Ψ⟩ =∑

jλ j

∣∣a j⟩

A

∣∣b j⟩

B , (5.21)

with real, positive Schmidt coefficients λ j, and∑

jλ2j = 1. This decom-

position is unique, and the sum runs at most to dA, the dimension ofthe smallest Hilbert space. Traditionally, we order the Schmidt coeffi-cients in descending order: λ1 ≥λ2 ≥ . . .. The total number of non-zeroλi is the Schmidt number.

The proof can be found in many graduate texts on quantum mechanics andquantum information theory.

Given the Schmidt decomposition for a bi-partite system, we can imme-diately write down the reduced density matrices for the sub-systems:

ρA =TrB(|Ψ⟩⟨Ψ|)=∑

jλ2

j∣∣a j

⟩A

⟨a j

∣∣ , (5.22)

and

ρB =TrA(|Ψ⟩⟨Ψ|)=∑

jλ2

j∣∣b j

⟩B

⟨b j

∣∣ . (5.23)

The basis states∣∣a j

⟩A and

∣∣b j⟩

B may have completely different physicalmeanings; here we care only that the states of the decomposition can belabelled with a single index, as opposed to two indices.

Conversely, when we have a single system in a mixed state

ρ =∑

jp j

∣∣a j⟩⟨

a j∣∣ , (5.24)

we can always construct a pure state |Ψ⟩ that obeys (λ j =pp j)

|Ψ⟩ =∑

jλ j

∣∣a j,b j⟩

, (5.25)

By virtue of the Schmidt decomposition. The state |Ψ⟩ is called the purifica-tion of ρ. Since many theorems are easier to prove for pure states than formixed states, purifications can make our work load significantly lighter.

58

When there is more than one non-zero λ j in Eq. (5.25), the state |Ψ⟩ isclearly entangled: there is no alternative choice of λ j due to the uniquenessof the Schmidt decomposition that would result in λ′

1 = 1 and all others zero.Moreover, the more uniform the values of λ j, the more the state is entangled.One possible measure for the amount of entanglement in |Ψ⟩ is the Shannonentropy H

H =−∑

jλ2

j log2λ2j . (5.26)

This is identical to the von Neumann entropy S of the reduced density ma-trix ρ of |Ψ⟩ given in Eq. (5.24):

S(ρ)=−Tr(ρ log2ρ) . (5.27)

Both entropies are measured in classical bits.How do we find the Schmidt decomposition? Consider the state |Ψ⟩ from

Eq. (5.20). The (not necessarily square) matrix C with elements c jk needsto be transformed into a single array of numbers λ j. This is achieved byapplying the singular-value decomposition:

c jk =∑

iu jidiivik , (5.28)

where u ji and vik are elements of unitary matrices U and V , respectively,and dii is a diagonal matrix with singular values λi. The vectors in theSchmidt decomposition become

|ai⟩ =∑

ju ji

∣∣φ j⟩

and |bi⟩ =∑k

vik∣∣ψk

⟩. (5.29)

This is probably a good time to remind ourselves about the singular-valuedecomposition. All we need to do is find U and V , the rest is just matrixmultiplication. To find U , we diagonalize CC† and find its eigenvectors.These form the columns of U . Similarly, we diagonalize C†C and arrangethe eigenvectors in columns to find V . If C is an n×m matrix, U should ben×n and V should be m×m.

59

5.3 Quantum teleportation

Probably the most extraordinary use of the quantum correlations present inentanglement is quantum teleportation. Alice and Bob share two entangledqubits, labelled 2 (held by Alice) and 3 (held by Bob), in the state (|0,0⟩23 +|1,1⟩23)/

p2. In addition, Alice holds a qubit in the state∣∣ψ⟩

1 =α |0⟩1 +β |1⟩1 . (5.30)

The object of quantum teleportation is to transfer the state of qubit 1 to qubit3, without either Alice or Bob gaining any information about α or β. To makethings extra hard, the three qubits must not change places (so Alice cannottake qubit 1 and bring it to Bob).

Classically, this is an impossible task: we cannot extract enough infor-mation about α and β with a single measurement to reproduce

∣∣ψ⟩faith-

fully, otherwise we could violate the no-cloning theorem. However, in quan-tum mechanics it can be done (without violating no-cloning). Write the totalstate as

∣∣χ⟩= ∣∣ψ⟩1

∣∣Φ+⟩23 =

1p2

(α |000⟩+α |011⟩+β |100⟩+β |111⟩) . (5.31)

Alice now performs a Bell measurement on her two qubits (1 and 2), whichproject them onto one of the Bell states

∣∣Φ±⟩12 or

∣∣Ψ±⟩12. We write |00⟩, |01⟩,

|10⟩, and |11⟩ in the Bell basis:

|00⟩ =∣∣Φ+⟩+|Φ−⟩

p2

|01⟩ =∣∣Ψ+⟩+|Ψ−⟩

p2

|10⟩ =∣∣Ψ+⟩−|Ψ−⟩

p2

|11⟩ =∣∣Φ+⟩−|Φ−⟩

p2

. (5.32)

We can use these substitutions to write the state∣∣χ⟩

before the measurement

60

as ∣∣χ⟩=12

[α

∣∣Φ+⟩12 |0⟩3 +α |Φ−⟩12 |0⟩3 +α

∣∣Ψ+⟩12 |1⟩3 +α |Ψ−⟩12 |1⟩3

+β∣∣Ψ+⟩

12 |0⟩3 −β |Ψ−⟩12 |0⟩3 +β∣∣Φ+⟩

12 |1⟩3 −β |Φ−⟩12 |1⟩3]

=12

[∣∣Φ+⟩(α |0⟩+β |1⟩)+|Φ−⟩ (α |0⟩−β |1⟩)

+∣∣Ψ+⟩

(β |0⟩+α |1⟩)+|Ψ−⟩ (β |0⟩−α |1⟩)] . (5.33)

Alice finds one of four possible outcomes:

Φ+ : Tr12(∣∣Φ+⟩⟨

Φ+∣∣ ∣∣χ⟩⟨χ∣∣) →

∣∣ψ⟩3 =α |0⟩+β |1⟩ ,

Φ− : Tr12(|Φ−⟩⟨Φ−|∣∣χ⟩⟨

χ∣∣) →

∣∣ψ⟩3 =α |0⟩−β |1⟩ ,

Ψ+ : Tr12(∣∣Ψ+⟩⟨

Ψ+∣∣ ∣∣χ⟩⟨χ∣∣) →

∣∣ψ⟩3 =α |1⟩+β |0⟩ ,

Ψ− : Tr12(|Ψ−⟩⟨Ψ−|∣∣χ⟩⟨

χ∣∣) →

∣∣ψ⟩3 =α |1⟩−β |0⟩ . (5.34)

From these outcomes, it is clear that the state held by Bob is different for thedifferent measurement outcomes of Alice’s Bell measurement. Let this sinkin for a moment: After setting up the entangled state between Alice and Bob,who may be literally light years apart, Bob has done absolutely nothing to hisqubit, yet its state is different depending on Alice’s measurement outcome.This suggests that there is some instantaneous communication taking place,possibly violating causality!

In order to turn the state of Bob’s qubit into the original state, Aliceneeds to send the measurement outcome to Bob. This will take two classicalbits, because there are four outcomes. The correction operators that Bobneed to apply are as follows:

Φ+ : I , Φ− : Z , Ψ+ : X , Ψ− : ZX . (5.35)

So in each case Bob needs to do something different to his qubit. To appreci-ate how remarkable this protocol is, here are some of its relevant properties:

1. No matter is transported, only the state of the system;

2. neither Alice nor Bob learns anything about α or β;

3. any attempt to use quantum teleportation for signaling faster thanlight is futile!

61

Exercises

1. Derive Eq. (5.15), and show that ρB = I/2 when Alice’s qubit is projectedonto |+⟩.

2. The state of a bipartite system is given by

|Ψ⟩ = 1p40

(|00⟩+3 |01⟩+2 |10⟩+5 |11⟩+ |12⟩) .

The first system is described by a two-dimensional Hilbert space, andthe second by a three-dimensional Hilbert space.

(a) Show that states with Schmidt number 1 are separable.

(b) Without doing elaborate calculations, what is the Schmidt num-ber of |Ψ⟩?

(c) Find the Schmidt decomposition of |Ψ⟩.

62

3. Quantum teleportation. Write the Bell states as∣∣ψnm⟩= (|0,0⊕n⟩+ (−1)m |1,1⊕n⟩) /

p2,

where ⊕ denotes addition modulo 2 and n,m = 0,1.

(a) Write∣∣ψnm

⟩in terms of

∣∣ψ00⟩

and the Pauli operators X and Zacting on the second qubit.

(b) Using the shared Bell state∣∣ψnm

⟩between Alice and Bob, and the

two-bit measurement outcome ( j,k) for Alice’s Bell measurement,determine the correction operator for Bob.

(c) We now generalize to N-dimensional systems. We define the N2

entangled states

∣∣ψnm⟩= 1p

N

N−1∑j=0

e2πi jn/N | j, j⊕m⟩ ,

where n⊕m = n+m mod N. Prove that this is an orthonormalbasis.

(d) Give the teleportation protocol for the N-dimensional Hilbert space.

(e) What is Bob’s state before he learns Alice’s measurement out-come?

4. Imperfect measurements.

(a) A two-qubit system (held by Alice and Bob) is in the anti-sym-metric Bell state |Ψ−⟩. Calculate the state of Bob’s qubit if Alicemeasures her qubit in the state |0⟩. Hint: write the measurementprocedure as a partial trace over Alice’s qubit.

(b) Now Alice’s measurement is imperfect, and when her apparatusindicates “0”, there was actually a small probability p that thequbit was projected onto the state |1⟩. What is Bob’s state?

(c) If Alice’s (imperfect) apparatus has only two measurement out-comes, what will Bob’s state be if she finds outcome “1”?

63

6 Evolution of Open Quantum Systems

We have considered mixed states, where the experimenter has incompleteinformation about the state preparation procedure, and we have also seenthat mixing arises in a system when it is entangled with another system.The combined system can still be pure, but the subsystem has become mixed.This phenomenon arises often when we want to describe systems that havesome interaction with their environment. The interaction creates entangle-ment, and the system taken by itself evolves from a pure state to a mixedstate. Such a system is called “open”, since it can leak quantum informationto the environment. The theory of open quantum systems revolves aroundthe so-called Lindblad equation.

6.1 The Lindblad equation

Next, we will derive the Lindblad equation, which is the direct extensionof the Heisenberg equation for the density operator, i.e., the mixed stateof a system. We have seen in Eq. (4.30) that formally, we can write theevolution of a density operator as a mathematical map E , such that thedensity operator ρ transforms into

ρ→ ρ′ = E (ρ)≡∑k

AkρA†k , (6.1)

where the Ak are the Kraus operators. Requiring that ρ′ is again a densityoperator (Tr(ρ′)= 1) leads to the restriction

∑k A†

k Ak = I.We want to describe an infinitesimal evolution of ρ, in order to give the

continuum evolution later on. We therefore have that

ρ′ = ρ+δρ =∑k

AkρA†k . (6.2)

Since δρ is very small, one of the Kraus operators must be close to the iden-tity. Without loss of generality we choose this to be A0, and then we canwrite

A0 = I+ (L0 − iK)δt and Ak = Lkpδt , (6.3)

where we introduced the Hermitian operators L0 and K , and the remainingLk are not necessarily Hermitian. We could have written A0 = I+L0δt and

64

keep L0 general (non-Hermitian as well), but it will be useful later on toexplicitly decompose it into Hermitian parts. We can now write

A0ρA†0 = ρ+

[(L0 − iK)ρ+ρ(L0 + iK)

]δt+O(δt2)

AkρA†k = LkρL†

kδt . (6.4)

We can substitute this into Eq. (6.2), to obtain up to first order in δt

δρ =[(

L0ρ+ρL0)− i(Kρ−ρK)+

∑k 6=0

LkρL†k

]δt . (6.5)

We now give the continuum evolution by dividing by δt and taking the limitδt → dt:

dρdt

=−i[K ,ρ]+ L0,ρ+∑k 6=0

LkρL†k , (6.6)

where A,B = AB+BA is the anti-commutator of A and B. We are almostthere, but we must determine what the different terms mean. Suppose weconsider the free evolution of the system. Eq. (6.6) must then reduce tothe Heisenberg equation for the density operator ρ in Eq. (4.7), and we seethat all Lk including L0 are zero, and K is proportional to the HamiltonianK = H/ħ. Again from the general property that Tr(ρ)= 1 we have

Tr(

dρdt

)= 0→ L0 =−1

2

∑k 6=0

L†kLk . (6.7)

This finally leads to the Lindblad equation

dρdt

= 1iħ [H,ρ]+ 1

2

∑k

(2LkρL†

k − L†kLk,ρ

). (6.8)

The operators Lk are chosen such that they model the relevant physicalprocesses. This may sound vague, but in practice it will be quite clear. Forexample, modelling a transition |1⟩→ |0⟩ without keeping track of where theenergy is going or coming from will require a single Lindblad operator

L = γ |0⟩⟨1| , (6.9)

where γ is a real parameter indicating the strength of the transition. Thiscan model both decay and excitations.

65

6.2 Positive and completely positive maps

We considered the evolution of the density operator under a family of Krausoperators in Eq. (4.30):

ρ→ ρ′ = E (ρ)=∑k

AkρA†k , (6.10)

where∑

k A†k Ak = I (that is, E is trace-preserving). When E transforms any

positive operator into another positive operator, we call it a positive map.We may be tempted to conclude that all positive maps correspond to physi-cally allowed transformations. After all, it maps density operators to densityoperators. Unfortunately, Nature (or Mathematics?) is not that tidy.

Consider the transpose of the density operator ρ → ρT , which acts ac-cording to

ρ =∑i jρ i j |i⟩⟨ j| → ρT =

∑i jρ ji |i⟩⟨ j| . (6.11)

You can verify immediately that the trace is preserved in this operation(check this!), and ρT is again a positive operator since the eigenvalues areidentical to those of ρ. For example, consider the qubit state (|0⟩+ i |1⟩)/

p2.

The density operator and its transpose are

ρ = 12

(1 −ii 1

)and ρT = 1

2

(1 i−i 1

). (6.12)

The transpose therefore corresponds to the state (|0⟩− i |1⟩)/p

2. Now con-sider that the qubit is part of an entangled state (|00⟩+|11⟩)/

p2. The density

operator is given by

ρ = 12

(|00⟩⟨00|+ |00⟩⟨11|+ |11⟩⟨00|+ |11⟩⟨11|) , (6.13)

and the partial transpose on the first qubit is

ρT = 12

(|00⟩⟨00|+ |10⟩⟨01|+ |01⟩⟨10|+ |11⟩⟨11|) . (6.14)

The eigenvalues of ρ are all positive, but ρT has a negative eigenvalue! SoρT cannot be a density operator. Consequently, it is not correct to say that

66

positive maps correspond to physical processes. We need to put anotherrestriction on maps.

From the example of the partial transpose, we can deduce that mapsmust not only be positive for the system S that they act on, but also pos-itive on larger systems that include S as a subsystem. When this is thecase, we call the map completely positive. There is a very important the-orem in mathematics, called Kraus’ Representation Theorem, which statesthat maps of the form in Eq. (6.10) with the restriction that

∑k A†

k Ak = I isa completely positive map, and moreover, that any completely positive mapcan be expressed in this form.

6.3 The Choi-Jamiołkowski isomorphism

There is a very interesting and useful relation between vectors and linearoperators in Hilbert space that goes by the name of the Choi-Jamiołkowskiisomorphism. Compare the operator

| j⟩⟨k| ∈H (6.15)

with the vector

| j⟩⊗ |k⟩ ∈H ⊗H . (6.16)

By turning the bra into a second ket we switch between Eq. (6.15) andEq. (6.16). The Choi-Jamiołkowski isomorphism says that the mathemat-ical structure of linear operators in H is exactly the same as that of thevectors in H ⊗H . This can be seen intuitively by viewing the operators| j⟩⟨k| as basis vectors in H with dimension d. For the case where d = 3, thebasis vectors can be written as nine matrices1 0 0

0 0 00 0 0

,

0 1 00 0 00 0 0

,

0 0 10 0 00 0 0

,

0 0 01 0 00 0 0

, . . .

0 0 00 0 00 0 1

.

You can see that we can construct any operator by multiplying these matri-ces with complex numbers and adding them. However, there is nothing spe-cial about the square arrangement, and we can place the zeros and ones ina vertical, vectorial arrangement. This is now a vector in a complex Hilbert

67

space of dimension d2. This isomorphism allows you to solve problems inoperator space that are hard in the vector space, and vice versa.

For a general operator U the isomorphism works as follows. Let

U =∑jk

U jk | j⟩⟨k| , (6.17)

for some basis | j⟩ in H . Applying the isomorphism, i.e., turning the bra intoa ket will define a new vector

|ΨU⟩ =∑jk

U jk | j⟩ |k⟩ . (6.18)

This will generally not be a normalised vector, but that does not concern ushere. We can also go the other way around, and define an operator given abi-partite vector:

|Ψ⟩ =∑jk

c jk | j⟩ |k⟩→UΨ =∑jk

c jk | j⟩⟨k| . (6.19)

In a way, this is all there is to the Choi-Jamiołkowski isomorphism.But what does it mean? We can compose two operations into a new op-

eration, but we can’t do this for vectors, so it is not clear where the isomor-phism takes us, physically speaking. To figure this out we will write theisomorphism in a slightly different way. Consider again the operator U :

U =∑jk

U jk | j⟩⟨k| →CJ

∑jk

U jk | j⟩ |k⟩ =∑

j| j⟩

(∑k

U jk |k⟩)

. (6.20)

However,∑

k U jk |k⟩ is just U | j⟩, and we can write this as

∑j| j⟩

(∑k

U jk |k⟩)= (IA ⊗UB)

∑j| j⟩A | j⟩B ≡ (IA ⊗UB)

∣∣Φ+⟩AB , (6.21)

where we defined |Φ+⟩ = ∑j | j, j⟩, and the subscript on U determines the

subspace it acts on. So the isomorphism in one direction can be summarisedby

U →CJ

|ΦU⟩ = (IA ⊗UB)∣∣Φ+⟩

, (6.22)

68

where the state |ΦU⟩ is isomorphic to U .The opposite direction of |ΦU⟩→U needs a little bit more work. We start

with an operator acting on |ψ⟩⟨ψ| and show that it equals

UA∣∣ψ⟩

A⟨ψ

∣∣U†A = BC

⟨Φ+∣∣(|ΦU⟩AB ⟨ΦU |⊗

∣∣ψ⟩C

⟨ψ

∣∣) ∣∣Φ+⟩BC . (6.23)

Since we need to take |ΦU⟩ to U , we start with the right-hand side (RHS)and rearrange so we end up with the left-hand side:

RHS= BC⟨Φ+∣∣(|ΦU⟩AB ⟨ΦU |⊗

∣∣ψ⟩C

⟨ψ

∣∣) ∣∣Φ+⟩BC

= BC⟨Φ+∣∣[(UA ⊗ IB)

∣∣Φ+⟩AB

⟨Φ+∣∣ (U†

A ⊗ IB)⊗∣∣ψ⟩

C⟨ψ

∣∣]∣∣Φ+⟩BC

=∑

jklmBC ⟨ j, j| (UA ⊗ IB)

∣∣k,k,ψ⟩

ABC⟨l, l,ψ

∣∣ (U†A ⊗ IB) |m,m⟩BC

=∑kl

C ⟨k|UA∣∣k,ψ

⟩AC

⟨l,ψ

∣∣U†A |l⟩C

=∑kl

UA |k⟩A (⟨k|ψ⟩)C ⟨l|U†A(⟨ψ|l⟩)C

=UA∣∣ψ⟩⟨

ψ∣∣U†

A . (6.24)

In the second line we used the isomorphism of Eq. (6.22) and in the third linewe expanded all |Φ+⟩. In the fourth line we evaluated the inner products ofvectors in the Hilbert space HB and used ⟨ j|k⟩ = δ jk. In the fifth line werecognise that the inner products (⟨k|ψ⟩)C are just the amplitudes of |ψ⟩ forbasis vector |k⟩, and the final result follows.

We can see from Eq. (6.23) that for the specific choice of U = I this is theteleportation protocol (up to a local correction). The state |ψ⟩ on system Cis transferred to system A, assuming that the measurement outcome of theBell measurement was |Φ+⟩. Using a nontrivial U then allows us to createthe resource state |ΦU⟩ for teleportation (instead of |Φ+⟩), and the result isthat we teleport the state |ψ⟩ with an added operation U . This is called gateteleportation. The name comes from quantum information theory, where Uis typically a quantum gate on a qubit or a set of qubits.

This trick is particularly useful when |ψ⟩ carries fragile quantum infor-mation5, and when applying U is difficult and noisy. Instead of applying U

5Which, when it is gone it cannot be retrieved thanks to the no-cloning theorem.

69

straight to |ψ⟩, and possibly destroy it, we can create the state |ΦU⟩. Sincethis is a known state (as opposed to |ψ⟩) we can try to create it as many timesas we want, until we get it right. Only then do we employ the gate teleporta-tion protocol. Eq. (6.23) then ensures that we obtain the state U |ψ⟩. This isthe fundamental principle behind measurement-based quantum computing,which has typically higher fault-tolerance thresholds than the circuit modelof quantum computing.

Exercises

1. Show that∑

k A†k Ak = I, prove that any non-Hermitian square matrix

can be written as A + iB, with A and B Hermitian, and prove thatL0 =−1

2∑

k 6=0 L†kLk.

2. Consider a two-level system (|0⟩, |1⟩) that has a dephasing process,modelled by the Lindblad operators L1 = γ |0⟩⟨1| and L2 = γ |1⟩⟨0|.

(a) write down the Lindblad equation (choose H = 0 for simplicity).

(b) Calculate the evolution of the pure states |0⟩ and |+⟩ at t = 0.Hint: write the density matrix in the Pauli matrix basis I, X ,Y , Z.What can you say about the equilibrium state of the system?

(c) Calculate and plot the entropy S(ρ) of the state ρ(t) as a functionof γ and t.

3. Calculate the eigenvalues of ρT in Eq. (6.14).

70

7 Orbital Angular Momentum and Spin

Angular momentum plays an important role in quantum mechanics, notonly as the orbital angular momentum of electrons orbiting the central po-tentials of nuclei, but also as the intrinsic magnetic moment of particles,known as spin, and even as isospin in high-energy particle physics.

7.1 Orbital angular momentum

From classical physics we know that the orbital angular momentum of aparticle is given by the cross product of its position and momentum

L= r×p or L i = εi jkr j pk , (7.1)

where we used Einstein’s summation convention for the indices. In quantummechanics, we can find the operator for orbital angular momentum by pro-moting the position and momentum observables to operators. The resultingorbital angular momentum operator turns out to be rather complicated, dueto a combination of the cross product and the fact that position and momen-tum do not commute. As a result, the components of orbital momentum donot commute with each other. When we use [r j, pk] = iħδ jk, the commuta-tion relation for the components of L becomes

[L i,L j]= iħεi jkLk . (7.2)

A set of relations like this is called an algebra, and the algebra here is calledclosed since we can take the commutator of any two elements L i and L j, andexpress it in terms of another element Lk. Another (simpler) closed algebrais [x, px]= iħI and [x, I]= [px, I]= 0.

Since the components of angular momentum do not commute, we cannotfind simultaneous eigenstates for Lx, L y, and Lz. We will choose one ofthem, traditionally denoted by Lz, and construct its eigenstates. It turnsout that there is another operator, a function of all L is, that commutes withany component L j, namely L2 = L2

x+L2y+L2

z. This operator is unique, in thatthere is no other operator that differs from L2 in a nontrivial way and stillcommutes with all L is. We can now construct simultaneous eigenvectors forLz and L2.

71

Since we are looking for simultaneous eigenvectors for the square of theangular momentum and the z-component, we expect that the eigenvectorswill be determined by two quantum numbers, l, and m. First, and withoutany prior knowledge, we can formally write down the eigenvalue equationfor Lz as

Lz |l,m⟩ = mħ|l,m⟩ , (7.3)

where m is some real number, and we included ħ to fit the dimensions of an-gular momentum. We will now proceed with the derivation of the eigenvalueequation for L2, and determine the possible values for l and m.

From the definition of L2, we have L2 −L2z = L2

x +L2y, and

⟨l,m|L2 −L2z |l,m⟩ = ⟨l,m|L2

x +L2y |l,m⟩ ≥ 0. (7.4)

The spectrum of Lz is therefore bounded by

l ≤ m ≤ l (7.5)

for some value of l. We derive the eigenvalues of L2 given these restrictions.First, we define the ladder operators

L± = Lx ± iL y with L− = L†+ . (7.6)

The commutation relations with Lz and L2 are

[Lz,L±]=±ħL± , [L+,L−]= 2ħLz , [L±,L2]= 0. (7.7)

Next, we calculate Lz(L+ |l,m⟩):

Lz(L+ |l,m⟩)= (L+Lz + [Lz,L+]) |l,m⟩ = mħL+ |l,m⟩+ħL+ |l,m⟩= (m+1)ħL+ |l,m⟩ . (7.8)

Therefore L+ |l,m⟩∝ |l,m+1⟩. By similar reasoning we find that L− |l,m⟩∝|l,m−1⟩. Since we already determined that −l ≤ m ≤ l, we must also requirethat

L+ |l, l⟩ = 0 and L− |l,−l⟩ = 0. (7.9)

72

Counting the states between −l and +l in steps of one, we find that thereare 2l +1 different eigenstates for Lz. Since 2l +1 is a positive integer, lmust be a half-integer (l = 0, 1

2 ,1, 32 ,2, . . .). Later we will restrict this further

to l = 0,1,2, . . .The next step towards finding the eigenvalues of L2 is to calculate the

following identity:

L−L+ = (Lx − iL y)(Lx + iL y)= L2x +L2

y + i[Lx,L y]=L2 −L2z −ħLz . (7.10)

We can then evaluate

L−L+ |l, l⟩ = 0 ⇒ (L2 −L2z −ħLz) |l, l⟩ =L2 |l, l⟩− (l2 + l)ħ2 |l, l⟩ = 0. (7.11)

It is left as an exercise (see exercise 1b) to show that

L2 |l,m⟩ = l(l+1)ħ2 |l,m⟩ . (7.12)

We now have derived the eigenvalues for Lz and L2.One aspect of our algebraic treatment of angular momentum we still

have to determine is the matrix elements of the ladder operators. We againuse the relation between L±, and Lz and L2:

⟨l,m|L−L+ |l,m⟩ =l∑

j=−l⟨l,m|L− |l, j⟩⟨l, j|L+ |l,m⟩ . (7.13)

Both sides can be rewritten as

⟨l,m|L2 −L2z −ħLz |l,m⟩ = ⟨l,m|L− |l,m+1⟩⟨l,m+1|L+ |l,m⟩ , (7.14)

where on the right-hand-side we used that only the m+1-term survives. Thisleads to

[l(l+1)−m(m+1)]ħ2 = |⟨l,m+1|L+ |l,m⟩ |2 . (7.15)

The ladder operators then act as

L+ |l,m⟩ = ħ√

l(l+1)−m(m+1) |l,m+1⟩ , (7.16)

73

and

L− |l,m⟩ = ħ√

l(l+1)−m(m−1) |l,m−1⟩ . (7.17)

We have seen that the angular momentum L is quantized, and that thisgives rise to a discrete state space parameterized by the quantum numbersl and m. However, we still have to restrict the values of l further, as men-tioned above. We cannot do this using only the algebraic approach (i.e., us-ing the commutation relations for L i), and we have to consider the spatialproperties of angular momentum. To this end, we write L i as

L i =−iħεi jk

(x j

∂

∂xk

), (7.18)

which follows directly from the promotion of r and p in Eq. (7.1) to quantummechanical operators. In spherical coordinates,

r =√

x2 + y2 + z2 , φ= arctan( y

x

), θ = arctan

(√x2 + y2

z

), (7.19)

the angular momentum operators can be written as

Lx =−iħ(−sinφ

∂

∂θ−cotθ cosφ

∂

∂φ

),

L y =−iħ(cosφ

∂

∂θ−cotθsinφ

∂

∂φ

),

Lz =−iħ ∂

∂φ,

L2 =−ħ2[

1sinθ

∂

∂θ

(sinθ

∂

∂θ

)+ 1

sin2θ

∂2

∂φ2

]. (7.20)

The eigenvalue equation for Lz then becomes

Lzψ(r,θ,φ)=−iħ ∂

∂φψ(r,θ,φ)= mħψ(r,θ,φ) (7.21)

We can solve this differential equation to find that

ψ(r,θ,φ)= ζ(r,θ) eimφ . (7.22)

74

A spatial rotation over 2π must return the wave function to its originalvalue, because ψ(r,θ,φ) must have a unique value at each point in space.This leads to ψ(r,θ,φ+2π)=ψ(r,θ,φ) and

eim(φ+2π) = eimφ , or e2πim = 1. (7.23)

This means that m is an integer, which in turn means that l must be aninteger also.

7.2 Spin

For orbital angular momentum we found that 2l+1 must be an integer, andmoreover the spatial properties of the wave function force l to be an integeras well. However, we can also construct states with half-integer l, but thismust then be an internal degree of freedom. This is called spin angularmomentum, or spin for short. We will show later in the course that the spinobservable is interpreted as an intrinsic magnetic moment of a system.

To describe spin, we switch from L to S, which is no longer related to rand p. The commutation relations between the components Si are the sameas for L i,

[Si,S j]= iħεi jkSk , (7.24)

so S and L obey the same algebra. The commutation relations between Sand L, r, and p vanish:

[Si,L j]= [Si, r j]= [Si, p j]= 0. (7.25)

Therefore, spin generates a whole new vector space, since it commutes withobservables that themselves do not commute (like r j and p j), and it is inde-pendent of the spatial degrees of freedom.

Since the commutation relations for S (its algebra) are the same as forL, we can immediately copy the algebraic structure of the eigenstates andeigenvalues:

Sz |s,ms⟩ = msħ|s,ms⟩ , with s = 0,12

,1,32

,2, . . .

S2 |s,ms⟩ = s(s+1)ħ2 |s,ms⟩ . (7.26)

75

When s = 12 , the system has two levels (a qubit) with spin eigenstates

∣∣12 , 1

2⟩

and∣∣1

2 ,−12⟩. We often write ms = +1

2 =↑ (“up”) and ms = −12 =↓ (“down”),

which finds its origin in the measurement outcomes of electron spin in aStern-Gerlach apparatus.

Now that we have introduced a whole new vector space related to spin,how do we write the wave function of a particle with spin? Without spin, thewave function is a normal single-valued function ψ(r, t) = ⟨r|ψ(t)⟩ of spaceand time coordinates. Now we have to add the spin degree of freedom. Foreach spin (↑ or ↓ when s = 1

2 ), we have a wave function ψ↑(r, t) for the particlewith spin up, and ψ↓(r, t) for the particle with spin down. We can write thisas a vector:

ψ(r, t)=(ψ↑(r, t)ψ↓(r, t)

). (7.27)

The spin degree of freedom generates a vector space, after all. The vector ψis called a spinor.

Expectation values are evaluated in the usual way, but now we have tosum over the spin degree of freedom, as well as integrate over space. Forexample, the probability of finding a particle with spin up in a region Ω ofspace is given by

p(↑,Ω)=∑

ms=↑,↓

∫Ω

drδms,↑|ψms (r, t)|2 =∫Ω

dr |ψ↑(r, t)|2 , (7.28)

and the expectation value of finding a particle with any spin in a region Ω ofspace is given by

p(Ω)=∑

ms=↑,↓

∫Ω

dr |ψms (r, t)|2 . (7.29)

The normalization of the spinor is such that

∑ms=↑,↓

∫V

dr |ψms (r, t)|2 = 1, (7.30)

where V is the entire space available to the particle (this may be the entireuniverse, or the volume of a box with impenetrable walls, etc.).

76

If spin is represented by (2s+1)-dimensional spinors (vectors), then spintransformations (operators) are represented by (2s+1)×(2s+1) matrices. Inthe two-dimensional case, we have by construction:

Sz |↑⟩ =ħ2

(10

)and Sz |↓⟩ =−ħ

2

(01

), (7.31)

which means that the matrix representation of Sz is given by

Sz =ħ2

(1 00 −1

). (7.32)

Next, the ladder operators act according to

S+ |↑⟩ = 0, S+ |↓⟩ = ħ|↑⟩ , S− |↑⟩ = ħ|↓⟩ , S− |↓⟩ = 0, (7.33)

which leads to the matrix representation

S+ =ħ(0 10 0

)and S− =ħ

(0 01 0

). (7.34)

From S± = Sx ± iSy we can then deduce that

Sx =ħ2

(0 11 0

)and Sy =

ħ2

(0 −ii 0

). (7.35)

We often define Si ≡ 12ħσi, where σi are the so-called Pauli matrices. Pre-

viously, we have called these matrices X , Y , and Z. The commutation rela-tions of the Pauli matrices are

[σi,σ j]= 2iεi jkσk or[σi

2,σ j

2

]= iεi jk

σk

2. (7.36)

Other important properties of the Pauli matrices are

σi,σ j≡σiσ j +σ jσi = 2δi jI (anti-commutator). (7.37)

They are both Hermitian and unitary, and the square of the Pauli matricesis the identity: σ2

i = I. Moreover, they obey an “orthogonality” relation

12

Tr(σiσ j)= δi j . (7.38)

77

The proof of this statement is as follows:

σiσ j =σiσ j +σ jσi −σ jσi = σi,σ j−σ jσi = 2δi jI−σ jσi . (7.39)

Taking the trace then yields

Tr(2δi jI−σ jσi)=Tr(σiσ j)=Tr(σ jσi) , (7.40)

or (using Tr(I)= 2)

2Tr(σ jσi)= 4δi j , (7.41)

which proves Eq. (7.38). If we define σ0 ≡ I, we can extend this proof to thefour-dimensional case

12

Tr(σµσν)= δµν , (7.42)

with µ,ν= 0,1,2,3.We can then write any 2×2 matrix as a sum over the two-dimensional

Pauli operators:

A =∑µ

aµσµ , (7.43)

since

12

Tr(Aσν)= 12

Tr

(∑µ

aµσµσν

)= 1

2

∑µ

aµTr(σµσν

)=∑µ

aµδµν = aν . (7.44)

The Pauli matrices and the identity matrix form a basis for the 2×2 matrices,and we can write

A = a0I+a ·σ =(

a0 +az ax − iayax + iay a0 −az

), (7.45)

where we used the notation σ = (σx,σy,σz).

78

L LL

S

S

S

J J

J

Figure 1: Addition of angular momentum.

1

Figure 3: Addition of angular momentum.

7.3 Total angular momentum

In general, a particle may have both spin and orbital angular momentum.Since L and S have the same dimensions, we can ask what is the total an-gular momentum J of the particle. We write this as

J=L+S≡L⊗ I+ I⊗S , (7.46)

which emphasizes that orbital and spin angular momentum are describedin distinct Hilbert spaces.

Since [L i,S j]= 0, we have

[Ji, J j]= [L i +Si,L j +S j]= [L i,L j]+ [Si,S j]

= iħεi jkLk + iħεi jkSk = iħεi jk(Lk +Sk)

= iħεi jk Jk . (7.47)

In other words, J obeys the same algebra as L and S, and we can imme-diately carry over the structure of the eigenvalues and eigenvectors from Land S.

In addition, L and S must be added as vectors. However, only one ofthe components of the total angular momentum can be sharp (i.e., having adefinite value). Recall that l and s are magnitudes of the orbital and spinangular momentum, respectively. We can determine the extremal valuesof J, denoted by ± j, by adding and subtracting the spin from the orbitalangular momentum, as shown in Figure 3:

|l− s| ≤ j ≤ l+ s . (7.48)

79

For example, when l = 1 and s = 12 , the possible values of j are j = 1

2 andj = 3

2 .The commuting operators for J are, first of all, J2 and Jz as we expect

from the algebra, but also the operators L2 and S2. You may think that Szand Lz also commute with these operators, but that it not the case:

[J2,Lz]= [(L+S)2,Lz]= [L2 +2L ·S+S2,Lz]= 2[L,Lz] ·S 6= 0. (7.49)

We can construct a full basis for total angular momentum in terms of J2 andJz, as before:

J2 ∣∣ j,m j⟩=ħ2 j( j+1)

∣∣ j,m j⟩

and Jz∣∣ j,m j

⟩= m jħ∣∣ j,m j

⟩. (7.50)

Alternatively, we can construct spin and orbital angular momentum eigen-states directly as a tensor product of the eigenstates

L2 |l,m⟩ |s,ms⟩ = ħ2l(l+1) |l,m⟩ |s,ms⟩ and Lz |l,m⟩ |s,ms⟩ = mħ|l,m⟩ |s,ms⟩ ,(7.51)

and

S2 |l,m⟩ |s,ms⟩ = ħ2s(s+1) |l,m⟩ |s,ms⟩ and Sz |l,m⟩ |s,ms⟩ = msħ|l,m⟩ |s,ms⟩ .(7.52)

Since the Lz and Sz do not commute with J2, the states∣∣ j,m j

⟩are not the

same as the states |l,m⟩ |s,ms⟩.

7.4 Composite systems with angular momentum

Now consider two systems, 1 and 2, with total angular momentum J1 andJ2, respectively. The total angular momentum is again additive, and givenby

J=J1 +J2 ≡J1 ⊗ I+ I⊗J2 . (7.53)

Completely analogous to the addition of spin and orbital angular momen-tum, we can construct the commuting operators J2, Jz, J2

1, and J22, but not

80

J1z and J2z. Again, we construct two natural bases for the total angularmomentum of the composite system, namely

| j,m⟩ or | j1,m1⟩⊗ | j2,m2⟩≡ | j1, j2,m1,m2⟩ . (7.54)

We want to know how the two bases relate to each other, because sometimeswe wish to talk about the angular momentum of the composite system, andat other times we are interested in the angular momentum of the subsys-tems. Since the second basis (as well as the first) in Eq. (7.54) forms a com-plete orthonormal basis, we can write

| j,m⟩ =∑

m1,m2

| j1, j2,m1,m2⟩⟨ j1, j2,m1,m2| j,m⟩ . (7.55)

The amplitudes ⟨ j1, j2,m1,m2| j,m⟩ are called Clebsch-Gordan coefficients,and we will now present a general procedure for calculating them.

Let’s first consider a simple example of two spin-12 systems, such as

two electrons. The spin basis for each electron is given by∣∣1

2 , 12⟩ = |↑⟩ and∣∣1

2 ,−12⟩= |↓⟩. The spin basis for the two electrons is therefore

| j1, j2,m1,m2⟩ ∈ |↑,↑⟩ , |↑,↓⟩ , |↓,↑⟩ , |↓,↓⟩ . (7.56)

The total spin is given by j = 12 + 1

2 = 1 and j = 12 − 1

2 = 0, so the four basisstates for total angular momentum are

| j,m⟩ ∈ |1,1⟩ , |1,0⟩ , |1,−1⟩ , |0,0⟩ . (7.57)

The latter state is the eigenstate for j = 0. The maximum total angularmomentum state |1,1⟩ can occur only when the two electron spins a parallel,and we therefore have

|1,1⟩ = |↑,↑⟩ =∣∣1

2 , 12⟩⊗ ∣∣1

2 , 12⟩

. (7.58)

To find the expansion of the other total angular momentum eigenstates interms of spin eigenstates we employ the following trick: use that J± = J1±+J2±. We can then apply J± to the state |1,1⟩, and J1± + J2± to the state∣∣12 , 1

2⟩⊗ ∣∣1

2 , 12⟩. This yields

J− |1,1⟩ = ħ√

j( j+1)−m(m−1) |1,0⟩ = ħp

2 |1,0⟩ . (7.59)

81

Similarly, we calculate

J1−∣∣1

2 , 12⟩=ħ

√12 (3

2 )− 12 (−1

2 )∣∣1

2 ,−12⟩=ħ

∣∣12 ,−1

2⟩

, (7.60)

and a similar result for J2−. Therefore, we find that

ħp

2 |1,0⟩ = ħ|↑,↓⟩+ħ|↓,↑⟩ =⇒ |1,0⟩ = |↑,↓⟩+ |↓,↑⟩p2

. (7.61)

Applying J− again yields

|1,−1⟩ = |↓,↓⟩ . (7.62)

This agrees with the construction of adding parallel spins. The three totalangular momentum states

|1,1⟩ = |↑,↑⟩ ,

|1,0⟩ = 1p2

(|↑,↓⟩+ |↓,↑⟩) ,

|1,−1⟩ = |↓,↓⟩ (7.63)

form a so-called triplet of states with j = 1. We now have to find the finalstate corresponding to j = 0, m = 0. The easiest way to find it at this pointis to require orthonormality of the four basis states, and this gives us thesinglet state

|0,0⟩ = 1p2

(|↑,↓⟩− |↓,↑⟩) . (7.64)

The singlet state has zero total angular momentum, and it is therefore in-variant under rotations.

In general, this procedure of finding the Clebsch-Gordan coefficients re-sults in multiplets of constant j. In the case of two spins, we have a tensorproduct of two two-dimensional spaces, which are decomposed in two sub-spaces of dimension 3 (the triplet) and 1 (the singlet), respectively. We writethis symbolically as

2⊗2=3⊕1 . (7.65)

82

If we had combined a spin 1 particle with a spin 12 particle, the largest multi-

plet would have been due to j = 1+ 12 = 3

2 , which is a 4-dimensional subspace,and the smallest subspace is due to j = 1− 1

2 = 12 , which is a two-dimensional

subspace:

3⊗2=4⊕2 . (7.66)

In general, the total angular momentum of two systems with angular mo-mentum k and l is decomposed into multiplets according to the followingrule (k ≥ l):

(2k+1)⊗(2l+1)= [2(k+ l)+1]⊕ [2(k+ l)−1]⊕ . . .⊕ [2(k− l)+1] ,(7.67)

or in terms of the dimensions of the subspaces (n ≥ m):

n⊗m= (n+m−1)⊕ (n+m−3)⊕ . . .⊕ (n−m+1) . (7.68)

Exercises

1. Angular momentum algebra.

(a) Prove the algebra given in Eq. (7.2). Also show that [L2,L i] = 0,and verify the commutation relations in Eq. (7.7).

(b) Show that L2 |l,m⟩ = l(l +1)ħ2 |l,m⟩. Use the fact that [L−,L2] =0.

2. Pauli matrices.

(a) Check that the matrix representation of the spin-12 operators obey

the commutation relations.

(b) Calculate the matrix representation of the Pauli matrices for s =1.

(c) Prove that exp[−iθ ·σ] is a 2×2 unitary matrix.

3. Isospin I describes certain particle families called multiplets, and thecomponents of the isospin obey the commutation relations [I i, I j] =iεi jkIk.

83

(a) What is the relation between spin and isospin?

(b) Organize the nucleons (proton and neutron), the pions (π+, π0,and π−), and the delta baryons (∆++, ∆+, ∆0, and ∆−) into multi-plets. You will have to determine their isospin quantum number.

(c) Give all possible decay channels of the delta baryons into pionsand nucleons (use charge and baryon number conservation).

(d) Calculate the relative decay ratios of ∆+ and ∆0 into the differentchannels.

4. A simple atom has orbital and spin angular momentum, and the Hamil-tonian for the atom contains a spin-orbit coupling term Hso = għL ·S,where għ is the coupling strength.

(a) Are orbital and spin angular momentum good quantum numbersfor this system? What about total angular momentum?

(b) Use first-order perturbation theory to calculate the energy shiftdue to the spin-orbit coupling term.

(c) Calculate the transition matrix elements of Hso in the basis|l,m; s,ms⟩.

5. Multiplets.

(a) A spin 32 particle and a spin 2 particle form a composite system.

How many multiplets are there, and what is the dimension of thelargest multiplet?

(b) How many multiplets do two systems with equal angular momen-tum have?

84

8 Identical Particles

We have so far looked at the quantum mechanical description of a few par-ticles with spin in the previous section, and particles that exhibit entan-glement in section 5. In all these cases, we assumed that the individualparticles could be distinguished from each other. For example, the two-electron state (|↑↓⟩− |↓↑⟩)/

p2 assumes that we have two electrons, one held

“over here”, and the other “over there”, and we can talk meaningfully abouttheir respective spins. The tensor product structure of our Hilbert space is amanifestation of our ability to label our particles unambiguously.

However, what happens when we place the two electrons inside a sealedbox? The wave functions of the electrons will quickly start to overlap. Sincethe electrons are identical particles, which according to basic quantum me-chanics do not have well-defined paths, we cannot keep track of which elec-tron is which inside the box. Not even in principle.

8.1 Symmetric and anti-symmetric states

The indistinguishability of identical particles means that we have to adjustour quantum mechanical description of these objects. There are two ways ofdoing this, namely via a modification of the allowed states and via a restruc-turing of the observables6. In this section we consider the restricted statespace, and in the next we will be considering the new observables.

First of all, since the total number of particles is an observable quantity(for example by measuring the total charge in the box), we can give theparticles an artificial labelling. The wave functions of the two particles arethen given by

∣∣ψ(r1)⟩

1 for particle 1 at position r1, and∣∣φ(r2)

⟩2 for particle

2 at position r2. Since we can swap the positions of the particle withoutobservable consequences, we find that there are two states that denote thesame physical situation:∣∣ψ(r1),φ(r2)

⟩12 and

∣∣ψ(r2),φ(r1)⟩

12 . (8.1)

However, we wish that each physically distinct situation has exactly onequantum state. Since there is no preference for either state, we can denote

6This is sometimes called second quantisation. This is a misnomer, since quantisationoccurs only once, when observables are promoted to operators.

85

the physical situation of identical particles at position r1 and r2 by the quan-tum state that is an equal weight over these two possibilities:

|Ψ(r1,r2)⟩12 =∣∣ψ(r1),φ(r2)

⟩12 + eiϕ ∣∣ψ(r2),φ(r1)

⟩12p

2. (8.2)

You can verify that swapping the positions r1 and r2 of the indistinguishableparticles incurs only a global (unobservable) phase. The question is now howwe should choose φ.

Suppose that the two identical particles in the box are electrons. Weknow From Pauli’s exclusion principle that the two electrons cannot be in thesame state. Therefore, when φ = ψ, the state in Eq. (8.2) should naturallydisappear:

|Ψ(r1,r2)⟩12 =∣∣ψ(r1),ψ(r2)

⟩12 + eiϕ ∣∣ψ(r2),ψ(r1)

⟩12p

2= 0, (8.3)

which means that for particles obeying Pauli’s exclusion principle we mustchoose eiϕ =−1. The quantum state of the two particles is anti-symmetric.

What about particles that do not obey Pauli’s exclusion principle? Thesemust be restricted to states that are orthogonal to the anti-symmetric states.In other words, they must be in states that are symmetric under the ex-change of two particles. For the two identical particles in a box, we thereforechoose the value eiϕ = +1, which makes the state orthogonal to the anti-symmetric state. The two possibilities for combining two identical particlesare therefore

|ΨS(r1,r2)⟩ =∣∣ψ(r1),φ(r2)

⟩+ ∣∣ψ(r2),φ(r1)⟩

p2

,

|ΨA(r1,r2)⟩ =∣∣ψ(r1),φ(r2)

⟩− ∣∣ψ(r2),φ(r1)⟩

p2

. (8.4)

These states include both the internal degrees of freedom, such as spin, andthe external degrees of freedom. So two electrons can still be in the state |↑↑⟩,as long as their spatial wave function is anti-symmetric. The particles thatare in a symmetric overall quantum state are bosons, while the particles inan overall anti-symmetric state are fermions.

86

We can extend this to N particles in a fairly straightforward manner. Forbosons, we sum over all possible permutations of r1 to rN :

|ΨS(r1, . . . ,rN )⟩ = 1pN!

∑perm(r1,...,rN )

∣∣ψ1(r1), . . . ,ψN (rN )⟩

. (8.5)

For fermions, the odd permutations pick up a relative minus sign:

|ΨA(r1, . . . ,rN )⟩ = 1pN!

∑even

∣∣ψ1(r1), . . . ,ψN (rN )⟩− 1p

N!

∑odd

∣∣ψ1(r1), . . . ,ψN (rN )⟩

.

(8.6)

This can be written compactly as the so-called Slater determinant

ΨA(r1, . . . ,rN )= 1pN!

∣∣∣∣∣∣∣∣∣∣ψ1(r1) ψ1(r2) . . . ψ1(rN )ψ2(r1) ψ2(r2) . . . ψ2(rN )

......

. . ....

ψN (r1) ψN (r2) . . . ψN (rN )

∣∣∣∣∣∣∣∣∣∣, (8.7)

where we removed the kets for notational convenience. The N particles inthe state |ΨA(r1, . . . ,rN )⟩ automatically obey the Pauli exclusion principle.

8.2 Creation and annihilation operators

The second, and particularly powerful way to implement the description ofidentical particles is via creation and annihilation operators. To see how thisdescription arises, consider some single-particle Hermitian operator A witheigenvalues a j. On physical grounds, and regardless of distinguishability,we require that n j particles in the eigenstate

∣∣a j⟩

of A must have a totalphysical value n j ×a j for the observable A. We can repeat this for all j, andobtain a potentially infinite set of basis vectors

|n1,n2,n3, . . .⟩ ,

for all integer values of n j, including zero. The spectrum of A can be boundedor unbounded, and discrete or continuous. It may even be degenerate. Forsimplicity, we consider here an unbounded, non-degenerate discrete spec-trum.

87

A special state is given by

|∅⟩ = |0,0,0, . . .⟩ , (8.8)

which indicates the state of no particles, or the vacuum. The numbers n jare called the occupation number, and any physical state can be written asa superposition of these states:

|Ψ⟩ =∞∑

n1,n2,n3,...=0cn1,n2,n3,... |n1,n2,n3, . . .⟩ . (8.9)

The basis states |n1,n2,n3, . . .⟩ span a linear vector space called a Fock spaceF . It is the direct sum of the Hilbert spaces for zero particles H0, oneparticle H1, two particles, etc.:

F =H0 ⊕H1 ⊕H2 ⊕H3 ⊕·· · (8.10)

Since |Ψ⟩ is now a superposition over different particle numbers, we re-quire operators that change the particle number. These are the creation andannihilation operators, a† and a respectively. Up to a proportionality con-stant that we will determine later, the action of these operators is definedby

a†j

∣∣n1,n2, . . . ,n j, . . .⟩∝ ∣∣n1,n2, . . . ,n j +1, . . .

⟩,

a j∣∣n1,n2, . . . ,n j, . . .

⟩∝ ∣∣n1,n2, . . . ,n j −1, . . .⟩

. (8.11)

These operators are each others’ Hermitian adjoint, since removing a parti-cle is the time reversal of adding a particle. Clearly, when an annihilationoperator attempts to remove particles that are not there, the result must bezero:

a j∣∣n1,n2, . . . ,n j = 0, . . .

⟩= 0. (8.12)

The vacuum is then defined as the state that gives zero when acted on byany annihilation operator: a j |∅⟩ = 0 for any j. Notice how we have so farsidestepped the problem of particle swapping; we exclusively used aspects ofthe total particle number.

88

So far, we have considered only the basis states of many particles for asingle observable A. What about other observables, in particular those thatdo not commute with A? We can make a similar construction. Suppose anobservable B has eigenvalues b j. We can construct creation and annihilationoperators b†

j and b j that act according to

b†j

∣∣m1,m2, . . . ,m j, . . .⟩∝ ∣∣m1,m2, . . . ,m j +1, . . .

⟩,

b j∣∣m1,m2, . . . ,m j, . . .

⟩∝ ∣∣m1,m2, . . . ,m j −1, . . .⟩

. (8.13)

where m j is the number of particles with value b j. Typically, the basis statesof two observables are related via a single unitary transformation

∣∣b j⟩ =

U∣∣a j

⟩for all j. How does this relate the creation and annihilation operators?

To answer this, let’s look at the single particle states. We can write thesingle-particle eigenstates

∣∣a j⟩

and∣∣b j

⟩as∣∣a j

⟩= a†j |∅⟩ and

∣∣b j⟩= b†

j |∅⟩ . (8.14)

We assume that U does not change the vacuum7, so U |∅⟩ = |∅⟩. This meansthat we can relate the two eigenstates via∣∣b j

⟩=U∣∣a j

⟩=Ua†j |∅⟩ =Ua†

j

(U†U

)|∅⟩ =Ua†

jU† |∅⟩ = b†

j |∅⟩ , (8.15)

where we have strategically inserted the identity U†U . This leads to theoperator transformation

b†j =Ua†

jU† . (8.16)

The Hermitian adjoint is easily calculated as b j = UaU†. It is left as anexercise for you to prove that

b†j =

∑k

u jk a†k and b j =

∑k

u∗k j ak , (8.17)

where u jk = ⟨ak|b j⟩.7This is a natural assumption when we are confined to the single particle Hilbert space,

but there are general unitary transformations for which this does not hold.

89

What are the basic properties of the creation and annihilation operators?In particular, we are interested in their commutation relations. We will nowderive these properties from what we have determined so far. First, notethat we can create two particles with eigenvalues bi and b j in the system inany order, and the only difference this can make is in the normalisation ofthe state:

b†i b†

j |Ψ⟩ =λb†j b

†i |Ψ⟩ , (8.18)

where λ is some complex number. Now let’s look at the same process in thebasis of the observable A. We have(

b†i b†

j −λb†j b

†i

)|Ψ⟩ =

∑kl

uik u jl

(a†

k a†l −λa†

l a†k

)|Ψ⟩ = 0. (8.19)

Since the uik and u jl are the matrix elements of an arbitrary unitary trans-formation, and the state |Ψ⟩ is certainly not zero, we require that

a†k a†

l −λa†l a

†k = 0. (8.20)

Since k and l are just dummy variables, we equally have

a†l a

†k −λa†

k a†l = 0. (8.21)

We now substitute Eq. (8.21) into Eq. (8.20) to eliminate a†l a

†k. This leads to

(1−λ2) a†k a†

l = 0, (8.22)

and therefore

λ=±1. (8.23)

The relation between different creation operators can thus take two forms.They can obey a commutation relation when λ=+1:

a†l a

†k − a†

k a†l =

[a†

l , a†k

]= 0, (8.24)

or they can obey an anti-commutation relation when λ=−1:

a†l a

†k + a†

k a†l =

a†

l , a†k

= 0, (8.25)

90

While creating the particles in different temporal order is not the same asswapping two particles, it should not come as a surprise that there are twopossible situations (the commutation relation and the anti-commutation re-lation). We encountered two possibilities in our previous approach as well,where we found that many-particle states are either symmetric or anti-symmetric. In fact, creation operators that obey the commutation rela-tion produce symmetric states, while creation operators that obey the anti-commutation relation produce anti-symmetric states. We also see that thecreation operators described by the anti-commutation relations naturallyobey Pauli’s exclusion principle. Suppose that we wish to create two iden-tical particles in the same eigenstate

∣∣a j⟩. The anti-commutation relations

say that a†j, a

†j= 0, so

a†2j = 0. (8.26)

Any higher powers of a†j will also be zero, and we can create at most one

particle in the state∣∣a j

⟩.

Taking the adjoint of the commutation relations for the creation opera-tors gives us the corresponding relations for the annihilation operators

al ak − ak al = [al , ak]= 0, (8.27)

or

al ak + ak al = al , ak= 0. (8.28)

The remaining question is now what the (anti-) commutation relations arefor products of creation and annihilation operators.

We proceed along similar lines as before. Consider the operators bi andb†

j with j 6= k, and apply them in different orders to a state |Ψ⟩.

bi b†j |Ψ⟩ =µ b†

j bi |Ψ⟩ . (8.29)

However, now we have to be a little bit careful about the case where wecreate and annihilate a particle in the same eigenstate, so we first considerthe case where i 6= j. In terms of the observable A, this becomes(

bi b†j −µ b†

j bi

)|Ψ⟩ =

∑kl

u∗ki u jl

(ak a†

l −µ a†l ak

)|Ψ⟩ = 0, (8.30)

91

and the same argumentation as before leads to µ=±1. For different k and lwe therefore find [

ak, a†l

]= 0 or

ak, a†

l

= 0. (8.31)

Let’s consider the special case where |Ψ⟩ = |∅⟩. We find(bi b

†j −µ b†

j bi

)|∅⟩ = bi b

†j |∅⟩ = δi j |∅⟩ . (8.32)

We want to know what effect this has on the case where l = k in Eq. (8.30):(bi b

†j −µ b†

j bi

)|∅⟩ =

∑k

u∗ki u jk

(ak a†

k −µ a†k ak

)|∅⟩ = δi j |∅⟩ . (8.33)

Since∑

k u∗ki u jk is just the component form of UU†, we find that

∑k u∗

ki u jk =δi j and the term in brackets must be equal to 1. This gives us our final setof relations: [

ak, a†k

]= 1 or

ak, a†

k

= 1. (8.34)

To summarise, we have two sets of algebras for the creation and annihilationoperators. The algebra in terms of the commutation relations is given by

[ak, al]=[a†

k, a†l

]= 0 and

[ak, a†

l

]= δkl . (8.35)

This algebra describes particles that obey Bose-Einstein statistics, or bosons.The algebra in terms of anti-commutation relations is given by

ak, al=a†

k, a†l

= 0 and

ak, a†

l

= δkl . (8.36)

This algebra describes particles that obey Fermi-Dirac statistics, or fermions.Finally, we have to determine the constant of proportionality for the cre-

ation and annihilation operators. We have already required that bi b†j |∅⟩ =

δi j |∅⟩. To determine the rest, we consider a new observable that gives usthe total number of particles in the system. We denote this observable by n,and we see that it must be additive over all particle numbers for the differenteigenvalues of A:

n =∑

jn j , (8.37)

92

where n j is the number of particles in the eigenstate∣∣a j

⟩. The total number

of particles does not change if we consider a different observable (althoughthe distribution typically will), so this relation is also true when we countthe particles in the states

∣∣b j⟩. Pretty much the only way we can achieve

this is to choose

n =∑

jn j =

∑j

a†j a j =

∑j

b†j b j . (8.38)

For the case of n j particles in state∣∣a j

⟩this gives

a†j a j

∣∣n j⟩= n j

∣∣n j⟩

, (8.39)

where we ignored the particles in other states |ak⟩ with k 6= j for brevity. Forthe Bose-Einstein case this leads to the relations

a j∣∣n j

⟩=√n j

∣∣n j −1⟩

and a†j

∣∣n j⟩=√

n j +1∣∣n j +1

⟩. (8.40)

For Fermi-Dirac statistics, the action of the creation and annihilation oper-ators on number states becomes

a j |0⟩ j = 0 and a†j |0⟩ j = e−iα |1⟩ j ,

a j |1⟩ j = eiα |0⟩ j and a†j |1⟩ j = 0. (8.41)

The phase factor eiα can be chosen ±1.

How do we construct operators using the creation and annihilation oper-ators? Suppose that a one-particle observable H has eigenvalues E j andeigenstates | j⟩. This can be, for example the Hamiltonian of the system,which ensures that the physical values of the particles (the eigenvalues) areadditive. The operator for many identical particles then becomes

H =∑

jE j n j =

∑j

E j a†j a j , (8.42)

which transforms according to Eq. (8.17). More generally, the operator maynot be written in the eigenbasis |n1,n2, . . .⟩, in which case it has the form

H =∑i j

Hi j b†i b j , (8.43)

93

where Hi j are matrix elements. The creation and annihilation operators a†j

and a j diagonalise H, and are sometimes called normal modes. The rea-son for this is that the creation and annihilation operators for bosons obeythe same mathematical rules as the raising and lowering operators for theharmonic oscillator. The index j then denotes different oscillators. A sys-tem of coupled oscillators can be decomposed into normal modes, which arethemselves isolated harmonic oscillators.

8.3 Bose-Einstein and Fermi-Dirac statistics

Finally, in this section we will derive the Bose-Einstein and Fermi-Diracstatistics. In particular, we are interested in the thermal equilibrium for alarge number of (non-interacting) identical particles with some energy spec-trum E j, which my be continuous.

Since the number of particles is not fixed, we are dealing with the GrandCanonical Ensemble. Its partition function Ξ is given by

Ξ=Tr[eµβn−βH

], (8.44)

where H is the many-body Hamiltonian, β = 1/kBT and µ is the chemicalpotential. The average number of particles with single particle energy E j isthen given by

⟨n j⟩ =−1β

∂ lnΞ∂E j

. (8.45)

For the simple case where H = ∑j E j n j and the creation and annihilation

operators obey the commutator algebra, the exponent can be written as

exp

[β

∑j

(µ−E j) a†j a j

]=

⊗j

∞∑n j=0

eβ(µ−E j)n j∣∣n j

⟩⟨n j

∣∣ , (8.46)

and the trace becomes

Ξ=∏

j

11− eβ(µ−E j)

. (8.47)

94

0E j

Xn j\

1E j

1Xn j\

Figure 4: Left: Bose-Einstein distribution for different temperatures (µ= 0).The lower the temperature, the more particles occupy the low energy states.Right: Fermi-Dirac distribution for different temperatures and µ = 1. Thefermions will not occupy energy states with numbers higher than 1, andtherefore higher energies are necessarily populated. The energy values E jform a continuum on the horizontal axis.

The average photon number for energy E j is

⟨n j⟩ =−1β

∂ lnΞ∂E j

=− 1βΞ

∂Ξ

∂E j= 1

e−β(µ−E j) −1. (8.48)

This is the Bose-Einstein distribution for particles with energy E j. It isshown for increasing E j in Fig. 5 on the left.

Alternatively, if the creation and annihilation operators obey the anti-commutation relations, the sum over n j in Eq. (8.48) runs not from 0 to ∞,but over 0 and 1. The partition function of the grand canonical ensemblethen becomes

Ξ=∏

j

[1+ eβ(µ−E j)

], (8.49)

and the average number of particles with energy E j becomes

⟨n j⟩ =− 1ħβΞ

∂Ξ

∂ω j= 1

e−β(µ−E j) +1. (8.50)

95

This is the Fermi-Dirac statistics for these particles, and it is shown in Fig. 5on the right. The chemical potential is the highest occupied energy at zerotemperature, and in solid state physics this is called the Fermi level. Notethe sign difference in the denominator with respect to the Bose-Einsteinstatistics.

Exercises

1. Calculate the Slater determinant for three electrons and show that notwo electrons can be in the same state.

2. Particle statistics.

(a) What is the probability of finding n bosons with energy E j in athermal state?

(b) What is the probability of finding n fermions with energy E j in athermal state?

3. Consider a system of (non-interacting) identical bosons with a dis-crete energy spectrum and a ground state energy E0. Furthermore,the chemical potential starts out lower than the ground state energyµ< E0.

(a) Calculate ⟨n0⟩ and increase the chemical potential to µ→ E0 (e.g.,by lowering the temperature). What happens when µ passes E0?

(b) What is the behaviour of ⟨nthermal⟩ ≡∑∞

j=1 ⟨n j⟩ as µ→ E0? Sketchboth ⟨n0⟩ and ⟨nthermal⟩ as a function of µ. What is the fraction ofparticles in the ground state at µ= E0?

(c) What physical process does this describe?

4. The process U = exp(r a†1a†

2 − r∗a1a2) with r ∈ C creates particles intwo systems, 1 and 2, when applied to the vacuum state |Ψ⟩ =U |∅⟩.

(a) Show that the bosonic operators a†1a†

1 and a1a2 obey the algebra

[K−,K+]= 2K0 and [K0,K±]=±K± ,

with K+ = K†−.

96

(b) For operators obeying the algebra in (a) we can write

erK+−r∗K− = exp[

r|r| tanh |r|K+

]exp

[−2ln(cosh |r|)K0

]×exp

[− r∗

|r| tanh |r|K−]

. (8.51)

Calculate the state |Ψ⟩ of the two systems.

(c) The amount of entanglement between two systems can be mea-sured by the entropy S(r) of the reduced density matrix ρ1 =Tr1[ρ] for one of the systems. Calculate S(r)=−Tr[ρ1 lnρ1].

(d) What is the probability of finding n particles in system 1? Com-pare this to your findings in question 2(a).

97

9 Many-Body Problems in Quantum Mechanics

In this final section we study a variety of problems in many-body quantummechanics. First, we introduce the Hartree-Fock method for taking into ac-count the effect of electron-electron interactions in atoms. Next, we describespin waves in magnetic materials using the Heisenberg model. Third, wedescribe the behaviour of an atom interacting with photons in a cavity, andintroduce the Jaynes-Cummings Hamiltonian. And finally, we take a brieflook at the basic ideas behind quantum field theory.

9.1 Interacting electrons in atomic shells

Previously, you have encountered the Schrödinger equation for a particlein a central potential, which can be interpreted as an electron bound to aproton in a hydrogen atom. We found that the energy levels are quantisedwith quantum number n. In addition, the spin and orbital angular momen-tum is quantised with quantum numbers ms, l, and m, respectively (weassume that s = 1

2 since we are considering electrons). We can denote theset of quantum numbers n, ms, l, and m by greek indices α, β,. . . , and thewave-functions uα(r) then form a complete orthonormal basis for the boundelectron.

It is tempting to keep this complete orthonormal basis for other atoms aswell, and assume that the ground state of an N-electron atom is the tensorproduct of the N lowest energy eigenstates, appropriately anti-symmetrizedvia the Slater determinant. Indeed, the periodic table is based on this as-sumption. However, this ignores the fact that the electrons interact witheach other, and the ground state of a many-electron atom is different. TheHartree-Fock method is designed to take this into account. It is a con-strained variational approach, in which the trial state that is optimised overis forced to be a Slater determinant in order to keep the correct particlestatistics. In this section we present the Hartree-Fock method, and arrive atthe Hartree-Fock equations, which can be solved iteratively. We follow thederivation given by Bransden and Joachain (Physics of Atoms and Molecules,1983 pp. 320-339).

First, we specify the Hamiltonian. Using the notation r i = |ri| for the dis-tance of the ith electron from the nucleus, and r i j = |ri −r j| for the distance

98

between electrons i and j, we find that

H = H1 +H2 =−N∑

i=1

( ħ2

2m∇2

i +Zr i

)+

N∑i< j=1

1r i j

, (9.1)

where we used units in which e/4πε0 = 1 and the Hamiltonian is dividedinto the single electron Hamiltonian (H1) and the inter-electron Hamilto-nian (H2). We choose as a normalised trial quantum state Ψ(r1, . . . ,rN ) theSlater determinant

Ψ(r1, . . . ,rN )= 1pN!

∣∣∣∣∣∣∣∣∣∣uα(r1) uβ(r1) . . . uν(r1)uα(r2) uβ(r2) . . . uν(r2)

......

. . ....

uα(rN ) uβ(rN ) . . . uν(rN )

∣∣∣∣∣∣∣∣∣∣, (9.2)

and calculate the expectation value of the Hamiltonian H. This must belarger or equal to the ground state E0

⟨Ψ|H|Ψ⟩ ≥ E0 , (9.3)

and varying the trial state then allows us to minimise the expectation value.This will get us close to the ground state energy.

Since the Slater determinant is a rather large expression, it will save usquite a bit of writing if we introduce the anti-symmetrisation operator A ,such that

Ψ(r1, . . . ,rN )=p

N! A uα(r1)uβ(r2) . . .uν(rN )≡p

N! AΦH , (9.4)

where ΦH(r1, . . . ,rN ) is the Hartree wave function. The operator A can thenbe written as a sum over permutations P of the labels α, β, . . .ν:

A = 1N!

∑P

(−1)P P , (9.5)

and A is a projection operator: A 2 = A = A †. Both H1 and H2 commutewith A .

99

Next, we calculate the expectation values ⟨Ψ|H1|Ψ⟩ and ⟨Ψ|H2|Ψ⟩. Since[H1,A ]= 0, we can write

⟨Ψ|H1|Ψ⟩ = N! ⟨ΦH|A H1A |ΦH⟩ = N! ⟨ΦH|H1A 2|ΦH⟩ = N! ⟨ΦH|H1A |ΦH⟩ .(9.6)

A permutation of the labels α, β, . . .ν leads to an orthonormal state and H1is a sum over one-electron Hamiltonians. We can therefore write this as

⟨Ψ|H1|Ψ⟩ = N! ⟨ΦH|H1A |ΦH⟩ =∑P

(−1)P⟨ΦH|H1 P|ΦH⟩ =N∑

i=1⟨ΦH|hi|ΦH⟩

=∑α

⟨uα(ri)|hi|uα(ri)⟩ ≡∑α

Iα

(9.7)

where hi is the single electron Hamiltonian

hi =− ħ2

2m∇2

i −Zr i

. (9.8)

Next, we calculate the expectation value ⟨Ψ|H2|Ψ⟩ of the two-electroninteraction Hamiltonians. Using the same reasoning as in Eq. (9.6), we findthat

⟨Ψ|H2|Ψ⟩ = N! ⟨ΦH|H2A |ΦH⟩ . (9.9)

Substituting the explicit form of A , we find

⟨Ψ|H2|Ψ⟩ =∑i< j

∑P

(−1)P⟨ΦH

∣∣∣∣ Pr i j

∣∣∣∣ΦH

⟩=

∑i< j

⟨ΦH

∣∣∣∣1−Pi j

r i j

∣∣∣∣ΦH

⟩, (9.10)

where Pi j is the exchange operator of electrons i and j. This expressionallows us to write

⟨Ψ|H2|Ψ⟩ =12

∑α,β

[⟨uα(ri)uβ(r j)

∣∣∣∣ 1r i j

∣∣∣∣uα(ri)uβ(r j)⟩

−⟨

uα(ri)uβ(r j)∣∣∣∣ 1r i j

∣∣∣∣uβ(ri)uα(r j)⟩]

. (9.11)

100

Note the swap of α and β in the last ket. This expectation value consists oftwo terms, namely the direct term

Jαβ ≡⟨

uα(ri)uβ(r j)∣∣∣∣ 1r i j

∣∣∣∣uα(ri)uβ(r j)⟩

, (9.12)

and the exchange term

Kαβ ≡⟨

uα(ri)uβ(r j)∣∣∣∣ 1r i j

∣∣∣∣uβ(ri)uα(r j)⟩

. (9.13)

The total expectation value therefore becomes

⟨Ψ|H|Ψ⟩ =∑α

Iα+12

∑α,β

(Jαβ−Kαβ

). (9.14)

The matrix elements Jαβ and Kαβ are real and symmetric in α and β.The second step towards the Hartree-Fock method is to find the mini-

mum of E ≡ ⟨Ψ|H|Ψ⟩ by varying the uα(ri). This means finding δE = 0.However, we must keep the functions uα(ri) orthonormal to each other, andthis imposes N2 constraints. We can incorporate these constraints in thevariational procedure by including Lagrange multipliers εαβ, and the varia-tional equation becomes

δE−∑α,β

εαβ δ⟨uα(r)|uβ(r)⟩ = 0. (9.15)

There is no explicit reference to electron positions ri in ⟨uα(r)|uβ(r)⟩ sincewe are only interested in its orthonormality properties. The Lagrange mul-tipliers εαβ form the elements of a Hermitian matrix.

The variational approach ultimately leads to a set of N coupled equa-tions:

Eα uα(ri)=hiuα(ri)+∑β

⟨uβ(r j)

∣∣ r−1i j

∣∣uβ(r j)⟩

uα(ri)

−∑β

⟨uβ(r j)

∣∣ r−1i j

∣∣uα(r j)⟩

uβ(ri) , (9.16)

which are known as the Hartree-Fock equations. These can be solved byiteration up to any desired precision.

101

9.2 Spin waves in solids

Consider a system of spins with a nearest-neighbour interaction. For a uni-form interaction in all direction, this is described by the Hamiltonian

H =±J∑(i, j)

Si ·S j =±J∑(i, j)

Sz,iSz, j +12

(S+,iS−, j +S−,iS+, j

). (9.17)

where J > 0 is the coupling strength between the spins,∑

(i, j) is the sumover all neighbouring pairs, and S± = Sx ± iSy. The physics described bythis Hamiltonian is known as the Heisenberg model. The sign of the cou-pling (here made explicit) determines whether the spins want to lign up inparallel (−J) or antiparallel (+J). The former situation describes ferromag-nets, while the later describes anti-ferromagnets. The spin operators fordifferent sites (i 6= j) commute with each other, while the spin operators atthe same site (i = j) obey the spin algebra of Eq. (7.24).

Both systems have a well-defined ground state. For the ferromagnet thisis the tensor product of the ground state of each individual spin. We are in-terested in the behaviour of the excitations with respect to this ground state.Due to the large degeneracy in the system (all the spins are of the samespecies with the same coupling J) the excitations act as identical quasi-particles. Consequently we can describe them using creation and annihi-lation operators. It turns out that they behave like bosons. You can thinkof an excitation as a higher spin value at some site that propagates to itsneighbours due to the interaction. This is called a spin wave.

Suppose that the spins are aligned in the positive z-direction (so we con-sider −J), and Sz has the maximum eigenvalue s. When the spin is loweredby ħ, this creates an excitation in the system, because the spin is no longerlined up. So the z-component of the spin at site j is given by

Sz, j = s− a†j a j , (9.18)

where a†j a j is the operator for the number of excitations at site j. Since S±

raise and lower the eigenvalue of Sz, we expect that S+ ∝ a and S− ∝ a†.When we insist on the commutation relation [S+,S−]= 2Sz, they become

S+, j =(2s− a†

j a j

) 12 a j and S−, j =

(2s− a†

j a j

) 12 a†

j . (9.19)

102

This is known as the Holstein-Primakoff transformation.For small numbers, the operators S± can be approximated as

S+, j 'p

2s a j and S−, j 'p

2s a†j . (9.20)

This allows us to write the Heisenberg Hamiltonian of Eq. (9.17) with −J tolowest order as

H =−J∑(i, j)

[s2 + s

(a†

i a j + ai a†j − a†

i ai − a†j a j

)]. (9.21)

For a simple cubic lattice of side L, lattice constant a and total number ofspins N = (L/a)3 we expect the spin waves to have wave vectors

k= 2πL

(m,n, o) with m,n, o ∈N , (9.22)

and 1 ≤ m,n, o ≤ L. The spin sites must now be indicated by a vector rinstead of a single number j, and the Fourier transformation of a†

r and ar isgiven by

ar =1pN

∑k

e−ik·r ak and a†r =

1pN

∑k

eik·r a†k , (9.23)

which transforms the Heisenberg Hamiltonian to

H =−3Js2N − JsN

∑r,d

∑k,k′

eir(k−k′)(eid·q −1

)a†

kak′

=−3Js2N − JsN

∑k′ε(k) a†

kak , (9.24)

where r is the position of a lattice site, d is the vector from a site to itsnearest neighbours, which takes care of the sum over nearest neighbours.This is a diagonal Hamiltonian with eigen-energies

ε(k)= 2Js (3−coskxa−coskya−coskza) . (9.25)

This is the dispersion relation for the spin waves, and to lowest order (cos x '1− 1

2 x2) it is quadratic:

ε(k)= Jsa2k2 . (9.26)

103

Spin waves are important when we want to manipulate magnetic propertieswith high frequency, such as in microwave devices. They carry energy, andare therefore a mechanism for dissipation.

For the case of anti-ferromagnets (+J), the ground state is harder tofind. Consider an anti-ferromagnet that is again a simple cubic lattice withalternating spin ±s and lattice constant a. We can think of this lattice as twosub-lattices with constant spin, and redefine the spins on the −s sub-latticeaccording to

Sx →−Sx , Sy → Sy , and Sz →−Sz . (9.27)

These operators still obey the commutation relations of spin (which S →−Swould not), and the Heisenberg Hamiltonian becomes

H =−J∑(i, j)

Sz,iSz, j +12

(S+,iS+, j +S−,iS−, j

). (9.28)

When we apply the Holstein-Primakoff transformation to this Hamiltonian,to first order we obtain

H =−J∑(i, j)

[s2 + s

(a†

i ai + b†j b j + ai b j + a†

i b†j

)], (9.29)

where a†i and ai are the creation and annihilation operators for the spin +s

sub-lattice, and b†j and b j are the creation and annihilation operators for the

original spin −s sub-lattice. After the Fourier transform of the creation andannihilation operators we get

H =−3Js2N +3Js∑k

[a†

kak + b†−kb−k + f (k)

(akb−k + a†

kb†−k

)], (9.30)

where f (k)= 13 (coskxa+coskya+coskza).

To find the ground state we must diagonalise H so that it is a sum overnumber operators. This will involve mixing creation and annihilation oper-ators. This is a unitary transformation that can be written as

ak = uk ck −vk d†−k and b−k = uk d−k −vk c†

k . (9.31)

104

This leads to the Hamiltonian

H =−3Js(s+1)N +∑kε(k)

(c†

k ck + d†−kd−k +1

), (9.32)

with the spin wave energy

ε(k)= 3Js(1− f (k)2) 1

2 . (9.33)

For small k the dispersion relation of the spin wave is linear in the wavevector, ε(k) '

p3Jsa k, which means that the spin waves behave markedly

different in ferromagnets and anti-ferromagnets.

9.3 An atom in a cavity

Consider a two-level atom with bare energy eigenstates |g⟩ and |e⟩ and en-ergy splitting ħω0. The free Hamiltonian H0 of the atom is given by

H0 =12ħω0

(1 00 −1

). (9.34)

The atom interacts with an electromagnetic wave, and the interaction isapproximately the coupling between the dipole moment d associated withthe |g⟩ ↔ |e⟩ transition and the electric field E, leading to the interactionHamiltonian

Hint =−d ·E , (9.35)

where d = −er, and E is the classical, complex-valued electric field at theposition of the atom. For an atom at the position r = 0, the electric field canbe written as

E= E0ε eiωt +E0ε∗e−iωt , (9.36)

where E0 is the real amplitude of the electric field ε is the polarsation vectorof the wave. The off-diagonal matrix elements of Hint are given by

⟨e|Hint|g⟩ = eE0⟨e|r|g⟩ ·ε eiωt +c.c., (9.37)

105

and c.c. denotes the complex conjugate. Since r has odd parity, the di-agonal matrix elements ⟨g|Hint|g⟩ and ⟨e|Hint|e⟩ vanish. When we definereg ≡ ⟨e|r|g⟩, the total Hamiltonian becomes

H =( 1

2ħω0 eE0r∗eg · (εeiωt +ε∗e−iωt)eE0reg · (εeiωt +ε∗e−iωt) −1

2ħω0

). (9.38)

Using the Rotating Wave Approximation (see exercise 9.1), this Hamiltoniancan be written as

H = ħ2

(ν Ω∗

Ω −ν), (9.39)

where we made the substitution

ν=ω−ω0 and Ω= 2eE0

ħ reg ·ε . (9.40)

We can use the standard matrix techniques in quantum mechanics to solvefor the eigenvalues, the eigenstates, and the time evolution of the atom.

Next, we consider the situation where atom is placed inside a cavity ofvolume V , and the electric field in the cavity has angular frequency ω withwave vector k propagating in the z-direction. Assume that the length of thecavity is a multiple of λ/2, such that ω is a resonant cavity mode. The fieldis very weak, so that the classical description of E is no longer sufficient. Inparticular, the field is made of photons, i.e., identical bosons. Consequently,we need to express E in terms of bosonic creation and annihilation operatorsa† and a, which create photons of frequency ω. Since the intensity of the fieldis proportional to both E 2

0 and a†a, we expect the electric field to be propor-tional to the creation and annihilation operators themselves. Hermiticityrequires that it is proportional to the sum of the creation and annihilationoperators. Furthermore, the electric field is a transverse standing wave cav-ity mode and must vanish at the mirrors due to the boundary conditionsimposed by Maxwell’s equations. The spatial amplitude variation thereforeincludes a factor sinkz. For linear polarisation the operator form of E thenbecomes

E(z, t)= ε√

ħωε0V

(a e−iωt + a† eiωt

)sinkz , (9.41)

106

where we assumed that ε is real8. The creation and annihilation operatorsare thus the amplitude operators of the field. Note that by the analogy withthe harmonic oscillator, the electric field operator acts as a position operatorof a particle in a harmonic potential well characterised by ω.

We again consider the dipole approximation of the atom in the field, andthe Hamiltonian is written as

Hint =−d · E= e r ·ε√

ħωε0V

sinkz(a e−iωt + a† eiωt

). (9.42)

The operator r can be written as

r= reg |e⟩⟨g|+r∗eg |g⟩⟨e| , (9.43)

and for notational simplicity, we define the coupling constant g as

g = ereg ·ε√

ħωε0V

sinkz . (9.44)

We again calculate the matrix elements of Hint as before, but this time wewrite the operator in terms of |g⟩⟨e| and |e⟩⟨g|:

Hint = g |e⟩⟨g|(a e−iωt + a† eiωt

)+ g∗ |g⟩⟨e|

(a e−iωt + a† eiωt

). (9.45)

It is convenient to express |g⟩⟨e| and |e⟩⟨g| in terms of the two-level raisingand lowering operators σ+ and σ−:

σ+ = |e⟩⟨g| and σ− = |g⟩⟨e| , (9.46)

with the commutation relation

[σ+,σ−]= 2σ3 with σ3 = |g⟩⟨g|− |e⟩⟨e| . (9.47)

The interaction Hamiltonian becomes

Hint = g σ+(a e−iωt + a† eiωt

)+ g∗ σ−

(a e−iωt + a† eiωt

). (9.48)

8This is true for linear polarisation. For elliptical polarisation ε will be complex. The sub-sequent derivation will be slightly modified (with more terms in Hint), but no extra technicalor conceptual difficulties arise.

107

Including the free Hamiltonian for the field and the two-level atom, thisbecomes in the Rotating Wave Approximation

HJC = 12ħω0 σ3 +ħω a†a+ g σ+ a + g∗σ− a† , (9.49)

This is the Jaynes-Cummings Hamiltonian for a two-level atom with energysplitting ħω0 interacting with a cavity mode of frequency ω. To achieve thestrongest coupling g, the volume of the cavity should be small, and the atomshould sit at an anti-node of the field.

Quantities are conserved when they commute with the Hamiltonian. Wecan identify two observables that satisfy this requirement, namely the num-ber of electrons

Pe = |g⟩⟨g|+ |e⟩⟨e| = I , (9.50)

and the total number of excitations

Ne = a†a+|e⟩⟨e| . (9.51)

This means that the Hamiltonian will not couple states with different totalexcitations.

In a real system, the cavity will not be perfect, and the excited state ofthe atom will suffer from spontaneous emission into modes other than thecavity mode. This can be modelled by a Lindblad equation for the joint stateρ of the atom and the cavity mode. The Lindblad operator for a leaky cavityis proportional to the annihilation operator a, with a constant of proportion-ality

pκ that denotes the leakage rate. The spontaneous emission of the

atom is modelled by the Lindblad operator pγσ−. The Lindblad equation

then becomes

dρdt

= 1iħ

[HJC,ρ

]+γσ−ρσ+− γ

2σ+σ−,ρ+κ aρa† − κ

2a†a,ρ . (9.52)

The research field of cavity quantum electrodynamics (or cavity QED) is de-voted in a large part to solving this equation.

108

9.4 Outlook: quantum field theory

We have surreptitiously introduced the basic elements of non-relativisticquantum field theory. Consider again the Heisenberg model, where we de-scribed a lattice of spins with nearest-neighbour interactions. If we take thelimit of the lattice constant a → 0 we end up with a continuum of creationand annihilation operators for each point in space. This is a field.

Traditionally we construct a quantum field theory from harmonic oscil-lators at each point in space. To this end, we characterise a classical har-monic oscillator with mass M in terms of its displacement q and velocity q.The equations of motion for the classical harmonic oscillator are the Euler-Lagrange equations

ddt

∂L∂q

− ∂L∂q

= 0, (9.53)

where L is the Lagrangian

L = 12

Mq2 − 12

K q2 , (9.54)

and K can be thought of as a spring constant. Substituting this L intoEq. (9.53) yields the familiar differential equations for the harmonic oscil-lator q+Ω2q = 0, with Ω2 = K /M.

Next, we arrange N particles in a one-dimensional lattice of length L andlattice constant a, where L = Na. Each particle’s displacement is coupled tothe displacement of it’s nearest neighbours by a spring with constant K . Theequations of motion of this set of coupled particles is given by

qn =Ω2 [(qn+1 − qn)+ (qn−1 − qn)] . (9.55)

We take the limit of a → 0 and N → ∞ while keeping L = Na fixed. Ourvariable qn(t) then becomes a field u(x, t), and it takes only a few lines ofalgebra to show that

u(x, t)= a2Ω2u′′(x, t)= v2u′′(x, t) with lima→0

aΩ= v . (9.56)

This is a wave equation, and v is the velocity of the wave. We can generalisethis immediately to three dimensions:

∂2u∂x2 + ∂2u

∂y2 + ∂2u∂z2 − 1

v2∂2u∂t2 = 0. (9.57)

109

In quantum field theory, we consider only these waves as the field excita-tions, and ignore the underlying particle structure we used to arrive at thisresult. We have already done something similar when we considered thespin waves in the Heisenberg model. Note also that the finite speed of propa-gation of waves means that we can make the theory Lorentz invariant whenv = c, the speed of light, and Eq. (9.57) becomes ∂µ∂µ u = 0.

The wave equation is typically derived from a Lagrangian L, or in thecase of a field theory, the Lagrangian density L . A massless scalar field isdescribed by the Lorentz-invariant Lagrangian density

L = 12

[(∂φ

∂x

)2+

(∂φ

∂x

)2+

(∂φ

∂x

)2− 1

c2

(∂φ

∂x

)2]= 1

2(∂µφ)(∂µφ) , (9.58)

where for technical reasons we redefined φ= u/p

a. The dispersion relationfor such a field is c2k2 =ω2, with k the wave number and ω the frequency ofthe wave. Similarly, a massive field is described by

L = 12

(∂µφ)(∂µφ)− 12

m2φ2 . (9.59)

The Euler-Lagrange equation for this Lagrangian density is the so-calledKlein-Gordon equation (

∂µ∂µ+ m2c2

ħ2

)φ= 0. (9.60)

The mass term leads to a new dispersion relation

c2ħ2k2 −ħ2ω2 +m2c4 = 0, (9.61)

and the group velocity for wave packets is

vg =dωdk

= c√1+µ2

with µ= mcħk

. (9.62)

For relativistic particles the momentum ħk is much larger than the restmass mc, and therefore vg approaches c.

We can solve the Klein-Gordon equation formally by writing

φ(r, t)=∫

dk√2(2π)3ωk

(a(k) eik·r−iωt +a∗(k) e−ik·r+iωt

), (9.63)

110

where ak is the complex amplitude of a wave with wave vector k. The fieldis essentially a superposition of (non-interacting) eigenmodes labelled by k,and we call this a free field. We can introduce interactions between the wavesin the field by adding higher-order terms to L with coupling constants ν,λ,. . .

L = 12

(∂µφ)(∂µφ)− 12

m2φ2 − ν

3!φ3 − λ

4!φ4 −·· · (9.64)

The solution to the Klein-Gordon equation is now no longer the correct solu-tion to the new equations of motion, but when the interaction is reasonablyweak we can use the solutions φ(r, t) of the free field as a starting point in aperturbation expansion.

So far, everything in this section has been a classical treatment. In orderto extend the theory to quantum mechanics we have to quantise the field.We achieve this by promoting the amplitudes in φ (and therefore φ itself) tooperators that obey commutation of anti-commutation relations

φ(r, t)=∫

dk√2(2π)3ωk

(a(k) eik·r−iωk t + a†(k) e−ik·r+iωk t

). (9.65)

These are the creation and annihilation operators for excitations of the field.For the Klein-Gordon equation they obey the bosonic commutation relations[

a(k), a(k′)]= [

a†(k), a†(k′)]= 0 and

[a(k), a†(k′)

]= δ3(k−k′) .

(9.66)

The state of the field can then be written as a superposition of Fock states.The field φ has now become an observable.

In quantum field theory, the excitations of the field are interpreted asparticles. All fundamental particles like quarks, electrons, photons, and theHiggs boson are excitations of a corresponding field. So the excitations ofthe Higgs field are Higgs bosons, and the excitations of the electromagneticfield are photons. Spin-1

2 particles obey the Dirac equation(iħγµ∂µ−mc

)ψ= 0, (9.67)

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where the γµ are 4×4 matrices

γ0 =(I 00 −I

), γ1 =

(0 σx

−σx 0

), γ2 =

(0 σy

−σy 0

), γ3 =

(0 σz

−σz 0

),

(9.68)

and ψ is a four-dimensional vector field called a spinor field. The solution tothe free Dirac field can be written as

ψ(r, t)=∑s

∫dk√

2(2π)3ωk

[bs(k) us(k) eik·r−iωk t + d†

s(k) vs(k) e−ik·r+iωk t]

,

(9.69)

where s =±12ħ is the spin value, and us(k) and vs(k) are two spinors carrying

the spin component of the field

us(k)=N

(χs

ħcσ·kħω+mc2 χs

)and vs(k)=N

(− ħcσ·kħω+mc2 χsχs

), (9.70)

where N is a normalisation constant, σ is a vector of Pauli matrices, and

χ+ =(10

)and χ− =

(01

). (9.71)

The creation and annihilation operators bs(k) and d†s(k) obey anti-commutation

relations:bs(k), br(k′)

= b†

s(k), b†r(k′)

= 0 and

bs(k), b†

r(k′)= δrs δ

3(k−k′) ,ds(k), dr(k′)

= d†

s(k), d†r(k′)

= 0 and

ds(k), d†

r(k′)= δrs δ

3(k−k′) ,

(9.72)bs(k), dr(k′)

= b†

s(k), d†r(k′)

= 0 and

bs(k), d†

r(k′)=

ds(k), b†

r(k′)= 0.

This means that the excitations of the Dirac field are fermions with spin 12 ,

such as the electron. You see immediately that ψ is not Hermitian due tothe appearance of d†. This means that ψ is not an observable and we cannotthink of the Dirac field as a quantised version of a classical observable field.There is no classical analog to the Dirac field. This is a consequence of the

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fact that the anti-particle of the Dirac excitations are not the same as theparticle itself. E.g., the positron is different from the electron. Anti-particlesare a quintessentially quantum mechanical phenomenon.

There is of course a lot more to quantum field theory than this. For exam-ple, the techniques for doing the perturbation expansion of interacting fieldsleads to Feynman diagrams, and renormalisation theory must be employedto deal with the infinities that crop up in the perturbation theory. Further-more, one has to choose the right Lagrangian density, and principles suchas gauge invariance and CPT invariance are imposed to constrain the possi-ble choices. This leads ultimately to the extraordinary successful StandardModel of particle physics. It the most fundamental theory of Nature that wehave, and it is tested to unprecedented accuracy.

Exercises

1. The Hamiltonian for a two-level atom in the presence of an electro-magnetic wave, as given in Eq. (9.38) depends on the time t. Thismakes it difficult to solve the Schrödinger equation, so in this exercisewe will get rid of the time dependence by applying the Rotating WaveApproximation.

(a) If∣∣ψ′(t)

⟩ = U(t)∣∣ψ(t)

⟩, find the Schrödinger equation for

∣∣ψ′(t)⟩.

What is the new Hamiltonian?

(b) Choose U(t) such that

U(t)=(eiωt/2 0

0 e−iωt/2

),

and H is given by Eq. (9.38). Show that the new Hamiltonian isgiven by Eq. (9.39).

2. A two-level atom is placed in a perfect cavity with an electromagneticfield of frequency ω.

(a) Show that the Jaynes-Cummings Hamiltonian can be written asa direct sum of 2×2 matrices Hn, and specify Hn.

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(b) Diagonalize Hn to find the energy values of the system, and cal-culate the eigenstates.

(c) At t = 0, the system is in the state∣∣ψ(0)

⟩ = |e,n⟩. Calculate thestate

∣∣ψ(t)⟩.

(d) Calculate the amount of entanglement between the atom and thecavity field. Use the relative entropy as the entanglement mea-sure.

(e) How long does the atom need to reside in the cavity in order toachieve maximum entanglement?

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